# Circles of given radius through two points

Circles of given radius through two points
You are encouraged to solve this task according to the task description, using any language you may know.

Given two points on a plane and a radius, usually two circles of given radius can be drawn through the points.

Exceptions
1. r==0.0 should be treated as never describing circles (except in the case where the points are coincident).
2. If the points are coincident then an infinite number of circles with the point on their circumference can be drawn, unless r==0.0 as well which then collapses the circles to a point.
3. If the points form a diameter then return two identical circles or return a single circle, according to which is the most natural mechanism for the implementation language.
4. If the points are too far apart then no circles can be drawn.

• Write a function/subroutine/method/... that takes two points and a radius and returns the two circles through those points, or some indication of special cases where two, possibly equal, circles cannot be returned.
• Show here the output for the following inputs:
```      p1                p2           r
0.1234, 0.9876    0.8765, 0.2345    2.0
0.0000, 2.0000    0.0000, 0.0000    1.0
0.1234, 0.9876    0.1234, 0.9876    2.0
0.1234, 0.9876    0.8765, 0.2345    0.5
0.1234, 0.9876    0.1234, 0.9876    0.0
```

## 11l

Translation of: Python
```T Circle((Float x, Float y, Float r))
F String()
R ‘Circle(x=#.6, y=#.6, r=#.6)’.format(.x, .y, .r)

T Error
String msg
F (msg)
.msg = msg

F circles_from_p1p2r(p1, p2, r) X(Error)
‘Following explanation at http://mathforum.org/library/drmath/view/53027.html’
I r == 0.0
V (x1, y1) = p1
V (x2, y2) = p2
I p1 == p2
X Error(‘coincident points gives infinite number of Circles’)
V (dx, dy) = (x2 - x1, y2 - y1)
V q = sqrt(dx ^ 2 + dy ^ 2)
I q > 2.0 * r
X Error(‘separation of points > diameter’)
V (x3, y3) = ((x1 + x2) / 2, (y1 + y2) / 2)
V d = sqrt(r ^ 2 - (q / 2) ^ 2)
V c1 = Circle(x' x3 - d * dy / q,
y' y3 + d * dx / q,
r' abs(r))
V c2 = Circle(x' x3 + d * dy / q,
y' y3 - d * dx / q,
r' abs(r))
R (c1, c2)

L(p1, p2, r) [((0.1234, 0.9876), (0.8765, 0.2345), 2.0),
((0.0000, 2.0000), (0.0000, 0.0000), 1.0),
((0.1234, 0.9876), (0.1234, 0.9876), 2.0),
((0.1234, 0.9876), (0.8765, 0.2345), 0.5),
((0.1234, 0.9876), (0.1234, 0.9876), 0.0)]
print("Through points:\n  #.,\n  #.\n  and radius #.6\nYou can construct the following circles:".format(p1, p2, r))
X.try
V (c1, c2) = circles_from_p1p2r(p1, p2, r)
print("  #.\n  #.\n".format(c1, c2))
X.handle Error v
print("  ERROR: #.\n".format(v.msg))```
Output:
```Through points:
(0.1234, 0.9876),
(0.8765, 0.2345)
You can construct the following circles:
Circle(x=1.863112, y=1.974212, r=2.000000)
Circle(x=-0.863212, y=-0.752112, r=2.000000)

Through points:
(0, 2),
(0, 0)
You can construct the following circles:
Circle(x=0.000000, y=1.000000, r=1.000000)
Circle(x=0.000000, y=1.000000, r=1.000000)

Through points:
(0.1234, 0.9876),
(0.1234, 0.9876)
You can construct the following circles:
ERROR: coincident points gives infinite number of Circles

Through points:
(0.1234, 0.9876),
(0.8765, 0.2345)
You can construct the following circles:
ERROR: separation of points > diameter

Through points:
(0.1234, 0.9876),
(0.1234, 0.9876)
You can construct the following circles:
```

## Action!

```INCLUDE "H6:REALMATH.ACT"

PROC Circles(CHAR ARRAY sx1,sy1,sx2,sy2,sr)
REAL x1,y1,x2,y2,r,x,y,bx,by,pb,cb,xx,yy
REAL two,tmp1,tmp2,tmp3

ValR(sx1,x1) ValR(sy1,y1)
ValR(sx2,x2) ValR(sy2,y2)
ValR(sr,r)   IntToReal(2,two)

Print("p1=(") PrintR(x1) Put(32)
PrintR(y1) Print(") p2=(")
PrintR(x2) Put(32) PrintR(y2)
Print(") r=") PrintR(r) Print(" -> ")

IF RealEqual(r,rzero) THEN
PrintE("Radius is zero, no circles") PutE()
RETURN
FI

RealSub(x2,x1,tmp1) ;tmp1=x2-x1
RealDiv(tmp1,two,x) ;x=(x2-x1)/2

RealSub(y2,y1,tmp1) ;tmp1=y2-y1
RealDiv(tmp1,two,y) ;y=(y2-y1)/2

RealMult(x,x,tmp1)      ;tmp1=x^2
RealMult(y,y,tmp2)      ;tmp2=y^2
Sqrt(tmp3,pb)           ;pb=sqrt(x^2+y^2)

IF RealEqual(pb,rzero) THEN
PrintE("Infinite circles")
ELSEIF RealGreater(pb,r) THEN
PrintE("Points are too far, no circles")
ELSE
RealMult(r,r,tmp1)      ;tmp1=r^2
RealMult(pb,pb,tmp2)    ;tmp2=pb^2
RealSub(tmp1,tmp2,tmp3) ;tmp3=r^2-pb^2
Sqrt(tmp3,cb)           ;cb=sqrt(r^2-pb^2)

RealMult(y,cb,tmp1) ;tmp1=y*cb
RealDiv(tmp1,pb,xx) ;xx=y*cb/pb

RealMult(x,cb,tmp1) ;tmp1=x*cb
RealDiv(tmp1,pb,yy) ;yy=x*cb/pb

RealSub(bx,xx,tmp1) ;tmp1=bx-xx
Print("c1=(") PrintR(tmp1) Put(32)

PrintR(tmp1) Print(") c2=(")

PrintR(tmp1) Put(32)

RealSub(by,yy,tmp1) ;tmp1=by-yy
PrintR(tmp1) PrintE(")")
FI
PutE()
RETURN

PROC Main()
Put(125) PutE() ;clear the screen
MathInit()
Circles("0.1234","0.9876","0.8765","0.2345","2.0")
Circles("0.0000","2.0000","0.0000","0.0000","1.0")
Circles("0.1234","0.9876","0.1234","0.9876","2.0")
Circles("0.1234","0.9876","0.8765","0.2345","0.5")
Circles("0.1234","0.9876","0.1234","0.9876","0.0")
RETURN```
Output:
```p1=(.1234 .9876) p2=(.8765 .2345) r=2 -> c1=(1.86311176 1.97421176) c2=(-0.86321176 -0.75211176)

p1=(0 2) p2=(0 0) r=1 -> c1=(0 1) c2=(0 1)

p1=(.1234 .9876) p2=(.1234 .9876) r=2 -> Infinite circles

p1=(.1234 .9876) p2=(.8765 .2345) r=.5 -> c1=(1.19528365 1.30638365) c2=(-0.1953836533 -0.0842836533)

p1=(.1234 .9876) p2=(.1234 .9876) r=0 -> Radius is zero, no circles
```

## ALGOL 68

Calculations based on the C solution.

```# represents a point                                                 #
MODE POINT = STRUCT( REAL x, REAL y );
# returns TRUE if p1 is the same point as p2, FALSE otherwise        #
OP = = ( POINT p1, POINT p2 )BOOL: x OF p1 = x OF p2 AND y OF p1 = y OF p2;

# represents a circle with centre c and radius r                     #
MODE CIRCLE = STRUCT( POINT c, REAL r );
# returns the difference in x-coordinate of two points               #
PRIO XDIFF = 5;
OP   XDIFF = ( POINT p1, POINT p2 )REAL: x OF p1 - x OF p2;
# returns the difference in y-coordinate of two points               #
PRIO YDIFF = 5;
OP   YDIFF = ( POINT p1, POINT p2 )REAL: y OF p1 - y OF p2;
# returns the distance between two points                            #
OP   -     = ( POINT p1, POINT p2 )REAL:
BEGIN
REAL x diff   = p1 XDIFF p2;
REAL y diff   = p1 YDIFF p2;
sqrt( ( x diff * xdiff ) + ( y diff * y diff ) )
END; # - #
# generate a human-readable version of the circle c                  #
OP TOSTRING = ( CIRCLE c )STRING:
+ fixed(      r OF c, -8, 4 )
+ " @("
+ fixed( x OF c OF c, -8, 4 )
+ ", "
+ fixed( y OF c OF c, -8, 4 )
+ ")"
);

# modes to represent the results of the circles procedure ...        #
# infinite number of circles                                         #
MODE INFINITECIRCLES = STRUCT( STRING t, REAL r );
# two possible circles                                               #
MODE TWOCIRCLES      = STRUCT( CIRCLE a, CIRCLE b );
# one possible circle results in a CIRCLE                            #
# no possible circles                                                #
MODE NOCIRCLES       = STRUCT( STRING reason, POINT p1, POINT p2, REAL r );
# mode returned by the circles procedure                             #
MODE POSSIBLECIRCLES = UNION( INFINITECIRCLES, TWOCIRCLES, CIRCLE, NOCIRCLES );

# returns the circles of radius r that can be drawn through          #
#         points p1 and p2                                           #
PROC circles = ( POINT p1, POINT p2, REAL r )POSSIBLECIRCLES:
IF r < 0 THEN # negative radius - there are no circles          #
NOCIRCLES( "negative radius", p1, p2, r )
ELIF p1 = p2 THEN # coincident points                           #
IF r = 0.0 THEN
# only one circle of radius 0 is possible               #
CIRCLE( p1, 0.0 )
ELSE
# an infinite number of circles can be drawn through    #
# the point                                             #
INFINITECIRCLES( "infinite", r )
FI
ELSE # two possible circles                                     #
REAL distance = p1 - p2;
IF   distance > 2 * r THEN
# the points are too far apart                          #
NOCIRCLES( "points too far apart", p1, p2, r )
ELIF distance = 2 * r THEN
# the points are on the diameter of the circle          #
CIRCLE( POINT( x OF p1 + ( ( p2 XDIFF p1 ) / 2 )
, y OF p1 + ( ( p2 YDIFF p1 ) / 2 )
)
, r
)
ELSE
# it is possible to draw two circles through the points #
REAL half x sum      = ( x OF p1 + x OF p2 ) / 2;
REAL half y sum      = ( y OF p1 + y OF p2 ) / 2;
REAL mirror distance = sqrt( ( r * r ) - ( ( distance * distance ) / 4 ) );
REAL x mirror        = ( mirror distance * ( y OF p1 - y OF p2 ) ) / distance;
REAL y mirror        = ( mirror distance * ( x OF p2 - x OF p1 ) ) / distance;
TWOCIRCLES( CIRCLE( POINT( half x sum + y mirror, half y sum + x mirror ), r )
, CIRCLE( POINT( half x sum - y mirror, half y sum - x mirror ), r )
)
FI
FI; # circles #

# test the circles procedure with the examples from the task    #

PROC print circles = ( REAL x1, y1, x2, y2, r )VOID:
BEGIN
CASE circles( POINT( x1, y1 ), POINT( x2, y2 ), r )
IN ( NOCIRCLES       n ): print( ( "No circles : ", reason OF n ) )
, ( TWOCIRCLES      t ): print( ( "Two circles: "
, TOSTRING a OF t
, ", "
, TOSTRING b OF t
)
)
, ( CIRCLE          c ): print( ( "One circle : ", TOSTRING c ) )
, ( INFINITECIRCLES i ): print( ( "Infinite circles" ) )
OUT BEGIN
print( ( "Unexpected circles result", newline ) );
stop
END
ESAC;
print( ( newline ) )
END; # print circles #
print circles( 0.1234, 0.9876,    0.8765, 0.2345,    2.0 );
print circles( 0.0000, 2.0000,    0.0000, 0.0000,    1.0 );
print circles( 0.1234, 0.9876,    0.1234, 0.9876,    2.0 );
print circles( 0.1234, 0.9876,    0.8765, 0.2345,    0.5 );
print circles( 0.1234, 0.9876,    0.1234, 0.9876,    0.0 )```
Output:
```Two circles: radius:  2.0000 @(  1.8631,   1.9742), radius:  2.0000 @( -0.8632,  -0.7521)
One circle : radius:  1.0000 @(  0.0000,   1.0000)
Infinite circles
No circles : points too far apart
One circle : radius:  0.0000 @(  0.1234,   0.9876)
```

## Arturo

```getPoint: function [p]->  ~{(x: |p\0|, y: |p\1|)}
getCircle: function [c]-> ~{(x: |c\0|, y: |c\1|, r: |c\2|)}

circles: function [p1, p2, r][
if r = 0 -> return "radius of zero"
if p1 = p2 -> return "coincident points gives infinite number of circles"

[dx, dy]: @[p2\0 - p1\0, p2\1 - p1\1]
if q > 2*r -> return "separation of points > diameter"

p3: @[(p1\0 + p2\0)/ 2, (p1\1 + p2\1) / 2]
d: sqrt (r*r) - (q/2)*(q/2)
return @[
@[(p3\0 - d*dy/q), (p3\1 + d*dx/q), abs r],
@[(p3\0 + d*dy/q), (p3\1 - d*dx/q), abs r]
]
]

loop [
[[0.1234, 0.9876], [0.8765, 0.2345], 2.0]
[[0.0000, 2.0000], [0.0000, 0.0000], 1.0]
[[0.1234, 0.9876], [0.1234, 0.9876], 2.0]
[[0.1234, 0.9876], [0.8765, 0.2345], 0.5]
[[0.1234, 0.9876], [0.1234, 0.9876], 0.0]
] 'tr [
[p1, p2, r]: tr
print ["Through points:\n  " getPoint p1 "\n  " getPoint p2]
print ["and radius" (to :string r)++"," "you can construct the following circles:"]
if? string? cic: <= circles p1 p2 r -> print ["   ERROR:" cic]
else [
[c1, c2]: cic
print ["  " getCircle c1]
print ["  " getCircle c2]
]
print ""
]
```
Output:
```Through points:
(x: 0.1234, y: 0.9876)
(x: 0.8764999999999999, y: 0.2345)
and radius 2.0, you can construct the following circles:
(x: 1.863111801658189, y: 1.974211801658189, r: 2.0)
(x: -0.8632118016581896, y: -0.7521118016581892, r: 2.0)

Through points:
(x: 0.0, y: 2.0)
(x: 0.0, y: 0.0)
and radius 1.0, you can construct the following circles:
(x: 0.0, y: 1.0, r: 1.0)
(x: 0.0, y: 1.0, r: 1.0)

Through points:
(x: 0.1234, y: 0.9876)
(x: 0.1234, y: 0.9876)
and radius 2.0, you can construct the following circles:
ERROR: coincident points gives infinite number of circles

Through points:
(x: 0.1234, y: 0.9876)
(x: 0.8764999999999999, y: 0.2345)
and radius 0.5, you can construct the following circles:
ERROR: separation of points > diameter

Through points:
(x: 0.1234, y: 0.9876)
(x: 0.1234, y: 0.9876)
and radius 0.0, you can construct the following circles:

## AutoHotkey

```CircleCenter(x1, y1, x2, y2, r){
d := sqrt((x2-x1)**2 + (y2-y1)**2)
x3 := (x1+x2)/2	, y3 := (y1+y2)/2
cx1 := x3 + sqrt(r**2-(d/2)**2)*(y1-y2)/d , 	cy1:= y3 + sqrt(r**2-(d/2)**2)*(x2-x1)/d
cx2 := x3 - sqrt(r**2-(d/2)**2)*(y1-y2)/d , 	cy2:= y3 - sqrt(r**2-(d/2)**2)*(x2-x1)/d
if (d = 0)
return "No circles can be drawn, points are identical"
if (d = r*2)
return "points are opposite ends of a diameter center = " cx1 "," cy1
if (d = r*2)
return "points are too far"
if (r <= 0)
if !(cx1 && cy1 && cx2 && cy2)
return "no solution"
return cx1 "," cy1 " & " cx2 "," cy2
}
```

Examples:

```data =
(
0.1234 0.9876 0.8765 0.2345 2.0
0.0000 2.0000 0.0000 0.0000 1.0
0.1234 0.9876 0.1234 0.9876 2.0
0.1234 0.9876 0.8765 0.2345 0.5
0.1234 0.9876 0.1234 0.9876 0.0
)

loop, parse, data, `n
{
obj := StrSplit(A_LoopField, " ")
MsgBox, % CircleCenter(obj[1], obj[2], obj[3], obj[4], obj[5])
}
```
Output:
```0.1234 0.9876 0.8765 0.2345 2.0 > 1.863112,1.974212 & -0.863212,-0.752112
0.0000 2.0000 0.0000 0.0000 1.0 > points are opposite ends of a diameter center = 0.000000,1.000000
0.1234 0.9876 0.1234 0.9876 2.0 > No circles can be drawn, points are identical
0.1234 0.9876 0.8765 0.2345 0.5 > no solution
0.1234 0.9876 0.1234 0.9876 0.0 > No circles can be drawn, points are identical
```

## AWK

```# syntax: GAWK -f CIRCLES_OF_GIVEN_RADIUS_THROUGH_TWO_POINTS.AWK
# converted from PL/I
BEGIN {
split("0.1234,0,0.1234,0.1234,0.1234",m1x,",")
split("0.9876,2,0.9876,0.9876,0.9876",m1y,",")
split("0.8765,0,0.1234,0.8765,0.1234",m2x,",")
split("0.2345,0,0.9876,0.2345,0.9876",m2y,",")
leng = split("2,1,2,0.5,0",r,",")
print("     x1      y1      x2      y2    r   cir1x   cir1y   cir2x   cir2y")
print("------- ------- ------- ------- ---- ------- ------- ------- -------")
for (i=1; i<=leng; i++) {
printf("%7.4f %7.4f %7.4f %7.4f %4.2f %s\n",m1x[i],m1y[i],m2x[i],m2y[i],r[i],main(m1x[i],m1y[i],m2x[i],m2y[i],r[i]))
}
exit(0)
}
function main(m1x,m1y,m2x,m2y,r,  bx,by,pb,x,x1,y,y1) {
if (r == 0) { return("radius of zero gives no circles") }
x = (m2x - m1x) / 2
y = (m2y - m1y) / 2
bx = m1x + x
by = m1y + y
pb = sqrt(x^2 + y^2)
if (pb == 0) { return("coincident points give infinite circles") }
if (pb > r) { return("points are too far apart for the given radius") }
cb = sqrt(r^2 - pb^2)
x1 = y * cb / pb
y1 = x * cb / pb
return(sprintf("%7.4f %7.4f %7.4f %7.4f",bx-x1,by+y1,bx+x1,by-y1))
}
```
Output:
```     x1      y1      x2      y2    r   cir1x   cir1y   cir2x   cir2y
------- ------- ------- ------- ---- ------- ------- ------- -------
0.1234  0.9876  0.8765  0.2345 2.00  1.8631  1.9742 -0.8632 -0.7521
0.0000  2.0000  0.0000  0.0000 1.00  0.0000  1.0000  0.0000  1.0000
0.1234  0.9876  0.1234  0.9876 2.00 coincident points give infinite circles
0.1234  0.9876  0.8765  0.2345 0.50 points are too far apart for the given radius
0.1234  0.9876  0.1234  0.9876 0.00 radius of zero gives no circles
```

## BASIC

### BASIC256

Translation of: Liberty BASIC
```function twoCircles(x1, y1, x2, y2, radio)
if x1 = x2 and y1 = y2 then  #Si los puntos coinciden
print "Los puntos son los mismos "
return ""
else
print "Hay cualquier número de círculos a través de un solo punto ("; x1; ", "; y1; ") de radio "; int(radio)
return ""
end if
end if
r2 = sqr((x1-x2)^2+(y1-y2)^2) / 2  #distancia media entre puntos
return ""
end if

#si no, calcular dos centros
cx = (x1+x2) / 2   #punto medio
cy = (y1+y2) / 2
#debe moverse desde el punto medio a lo largo de la perpendicular en dd2
dd2 = sqr(radio^2 - r2^2)   #distancia perpendicular
dx1 = x2-cx           #vector al punto medio
dy1 = y2-cy
dx = 0-dy1 / r2*dd2   #perpendicular:
dy = dx1 / r2*dd2     #rotar y escalar
print " -> Circulo 1 ("; cx+dy; ", "; cy+dx; ")"     #dos puntos, con (+)
print " -> Circulo 2 ("; cx-dy; ", "; cy-dx; ")"     #y (-)
return ""
end function

x1 = 0.1234 : y1 = 0.9876 : x2 = 0.8765 : y2 = 0.2345 : radio = 2.0
print "Puntos "; "("; x1; ","; y1; "), ("; x2; ","; y2; ")"; ", Radio "; int(radio)
print twoCircles (x1, y1, x2, y2, radio)
x1 = 0.0000 : y1 = 2.0000 : x2 = 0.0000 : y2 = 0.0000 : radio = 1.0
print "Puntos "; "("; x1; ","; y1; "), ("; x2; ","; y2; ")"; ", Radio "; int(radio)
print twoCircles (x1, y1, x2, y2, radio)
x1 = 0.1234 : y1 = 0.9876 : x2 = 0.12345 : y2 = 0.9876 : radio = 2.0
print "Puntos "; "("; x1; ","; y1; "), ("; x2; ","; y2; ")"; ", Radio "; int(radio)
print twoCircles (x1, y1, x2, y2, radio)
x1 = 0.1234 : y1 = 0.9876 : x2 = 0.8765 : y2 = 0.2345 : radio = 0.5
print "Puntos "; "("; x1; ","; y1; "), ("; x2; ","; y2; ")"; ", Radio "; int(radio)
print twoCircles (x1, y1, x2, y2, radio)
x1 = 0.1234 : y1 = 0.9876 : x2 = 1234 : y2 = 0.9876 : radio = 0.0
print "Puntos "; "("; x1; ","; y1; "), ("; x2; ","; y2; ")"; ", Radio "; int(radio)
print twoCircles (x1, y1, x2, y2, radio)
end```

## C

```#include<stdio.h>
#include<math.h>

typedef struct{
double x,y;
}point;

double distance(point p1,point p2)
{
return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}

{
double separation = distance(p1,p2),mirrorDistance;

if(separation == 0.0)
{
radius == 0.0 ? printf("\nNo circles can be drawn through (%.4f,%.4f)",p1.x,p1.y):
printf("\nInfinitely many circles can be drawn through (%.4f,%.4f)",p1.x,p1.y);
}

{
printf("\nGiven points are opposite ends of a diameter of the circle with center (%.4f,%.4f) and radius %.4f",(p1.x+p2.x)/2,(p1.y+p2.y)/2,radius);
}

{
printf("\nGiven points are farther away from each other than a diameter of a circle with radius %.4f",radius);
}

else
{

printf("\nTwo circles are possible.");
printf("\nCircle C1 with center (%.4f,%.4f), radius %.4f and Circle C2 with center (%.4f,%.4f), radius %.4f",(p1.x+p2.x)/2 + mirrorDistance*(p1.y-p2.y)/separation,(p1.y+p2.y)/2 + mirrorDistance*(p2.x-p1.x)/separation,radius,(p1.x+p2.x)/2 - mirrorDistance*(p1.y-p2.y)/separation,(p1.y+p2.y)/2 - mirrorDistance*(p2.x-p1.x)/separation,radius);
}
}

int main()
{
int i;

point cases[] =
{	{0.1234, 0.9876},    {0.8765, 0.2345},
{0.0000, 2.0000},    {0.0000, 0.0000},
{0.1234, 0.9876},    {0.1234, 0.9876},
{0.1234, 0.9876},    {0.8765, 0.2345},
{0.1234, 0.9876},    {0.1234, 0.9876}
};

for(i=0;i<5;i++)
{
printf("\nCase %d)",i+1);
}

return 0;
}
```
test run:
```Case 1)
Two circles are possible.
Circle C1 with center (1.8631,1.9742), radius 2.0000 and Circle C2 with center (-0.8632,-0.7521), radius 2.0000
Case 2)
Given points are opposite ends of a diameter of the circle with center (0.0000,1.0000) and radius 1.0000
Case 3)
Infinitely many circles can be drawn through (0.1234,0.9876)
Case 4)
Given points are farther away from each other than a diameter of a circle with radius 0.5000
Case 5)
No circles can be drawn through (0.1234,0.9876)
```

## C#

Works with: C sharp version 6
```using System;
{
public static void Main()
{
double[][] values = new double[][] {
new [] { 0.1234, 0.9876, 0.8765, 0.2345,   2 },
new [] { 0.0,       2.0,    0.0,    0.0,   1 },
new [] { 0.1234, 0.9876, 0.1234, 0.9876,   2 },
new [] { 0.1234, 0.9876, 0.8765, 0.2345, 0.5 },
new [] { 0.1234, 0.9876, 0.1234, 0.9876,   0 }
};

foreach (var a in values) {
var p = new Point(a[0], a[1]);
var q = new Point(a[2], a[3]);
Console.WriteLine(\$"Points {p} and {q} with radius {a[4]}:");
try {
var centers = FindCircles(p, q, a[4]);
Console.WriteLine("\t" + string.Join(" and ", centers));
} catch (Exception ex) {
Console.WriteLine("\t" + ex.Message);
}
}
}

static Point[] FindCircles(Point p, Point q, double radius) {
if(p == q) return new [] { p };
else throw new InvalidOperationException("No circles.");
}
if (p == q) throw new InvalidOperationException("Infinite number of circles.");

double sqDistance = Point.SquaredDistance(p, q);
if (sqDistance > sqDiameter) throw new InvalidOperationException("Points are too far apart.");

Point midPoint = new Point((p.X + q.X) / 2, (p.Y + q.Y) / 2);
if (sqDistance == sqDiameter) return new [] { midPoint };

double distance = Math.Sqrt(sqDistance);
double ox = d * (q.X - p.X) / distance, oy = d * (q.Y - p.Y) / distance;
return new [] {
new Point(midPoint.X - oy, midPoint.Y + ox),
new Point(midPoint.X + oy, midPoint.Y - ox)
};
}

public struct Point
{
public Point(double x, double y) : this() {
X = x;
Y = y;
}

public double X { get; }
public double Y { get; }

public static bool operator ==(Point p, Point q) => p.X == q.X && p.Y == q.Y;
public static bool operator !=(Point p, Point q) => p.X != q.X || p.Y != q.Y;

public static double SquaredDistance(Point p, Point q) {
double dx = q.X - p.X, dy = q.Y - p.Y;
return dx * dx + dy * dy;
}

public override string ToString() => \$"({X}, {Y})";

}
}
```
Output:
```Points (0.1234, 0.9876) and (0.8765, 0.2345) with radius 2:
(1.86311180165819, 1.97421180165819) and (-0.86321180165819, -0.752111801658189)
Points (0, 2) and (0, 0) with radius 1:
(0, 1)
Points (0.1234, 0.9876) and (0.1234, 0.9876) with radius 2:
Infinite number of circles.
Points (0.1234, 0.9876) and (0.8765, 0.2345) with radius 0.5:
Points are too far apart.
Points (0.1234, 0.9876) and (0.1234, 0.9876) with radius 0:
(0.1234, 0.9876)```

## C++

Works with: C++11
```#include <iostream>
#include <cmath>
#include <tuple>

struct point { double x, y; };

bool operator==(const point& lhs, const point& rhs)
{ return std::tie(lhs.x, lhs.y) == std::tie(rhs.x, rhs.y); }

enum result_category { NONE, ONE_COINCEDENT, ONE_DIAMETER, TWO, INFINITE };

using result_t = std::tuple<result_category, point, point>;

double distance(point l, point r)
{ return std::hypot(l.x - r.x, l.y - r.y); }

result_t find_circles(point p1, point p2, double r)
{
point ans1 { 1/0., 1/0.}, ans2 { 1/0., 1/0.};
if (p1 == p2) {
if(r == 0.) return std::make_tuple(ONE_COINCEDENT, p1,   p2  );
else        return std::make_tuple(INFINITE,       ans1, ans2);
}
point center { p1.x/2 + p2.x/2, p1.y/2 + p2.y/2};
double half_distance = distance(center, p1);
if(half_distance > r)      return std::make_tuple(NONE,         ans1,   ans2);
if(half_distance - r == 0) return std::make_tuple(ONE_DIAMETER, center, ans2);
double root = sqrt(pow(r, 2.l) - pow(half_distance, 2.l)) / distance(p1, p2);
ans1.x = center.x + root * (p1.y - p2.y);
ans1.y = center.y + root * (p2.x - p1.x);
ans2.x = center.x - root * (p1.y - p2.y);
ans2.y = center.y - root * (p2.x - p1.x);
return std::make_tuple(TWO, ans1, ans2);
}

void print(result_t result, std::ostream& out = std::cout)
{
point r1, r2; result_category res;
std::tie(res, r1, r2) = result;
switch(res) {
case NONE:
out << "There are no solutions, points are too far away\n"; break;
case ONE_COINCEDENT: case ONE_DIAMETER:
out << "Only one solution: " << r1.x << ' ' << r1.y << '\n'; break;
case INFINITE:
out << "Infinitely many circles can be drawn\n"; break;
case TWO:
out << "Two solutions: " << r1.x << ' ' << r1.y << " and " << r2.x << ' ' << r2.y << '\n'; break;
}
}

int main()
{
constexpr int size = 5;
const point points[size*2] = {
{0.1234, 0.9876}, {0.8765, 0.2345}, {0.0000, 2.0000}, {0.0000, 0.0000},
{0.1234, 0.9876}, {0.1234, 0.9876}, {0.1234, 0.9876}, {0.8765, 0.2345},
{0.1234, 0.9876}, {0.1234, 0.9876}
};
const double radius[size] = {2., 1., 2., .5, 0.};

for(int i = 0; i < size; ++i)
}
```
Output:
```Two solutions: 1.86311 1.97421 and -0.863212 -0.752112
Only one solution: 0 1
Infinitely many circles can be drawn
There are no solutions, points are too far away
Only one solution: 0.1234 0.9876
```

## D

Translation of: Python
```import std.stdio, std.typecons, std.math;

class ValueException : Exception {
this(string msg_) pure { super(msg_); }
}

struct V2 { double x, y; }
struct Circle { double x, y, r; }

/**Following explanation at:
http://mathforum.org/library/drmath/view/53027.html
*/
Tuple!(Circle, Circle)
circlesFromTwoPointsAndRadius(in V2 p1, in V2 p2, in double r)
pure in {
assert(r >= 0, "radius can't be negative");
} body {
enum nBits = 40;

if (r.abs < (1.0 / (2.0 ^^ nBits)))

if (feqrel(p1.x, p2.x) >= nBits &&
feqrel(p1.y, p2.y) >= nBits)
throw new ValueException("coincident points give" ~
" infinite number of Circles");

// Delta between points.
immutable d = V2(p2.x - p1.x, p2.y - p1.y);

// Distance between points.
immutable q = sqrt(d.x ^^ 2 + d.y ^^ 2);
if (q > 2.0 * r)
throw new ValueException("separation of points > diameter");

// Halfway point.
immutable h = V2((p1.x + p2.x) / 2, (p1.y + p2.y) / 2);

// Distance along the mirror line.
immutable dm = sqrt(r ^^ 2 - (q / 2) ^^ 2);

return typeof(return)(
Circle(h.x - dm * d.y / q, h.y + dm * d.x / q, r.abs),
Circle(h.x + dm * d.y / q, h.y - dm * d.x / q, r.abs));
}

void main() {
foreach (immutable t; [
tuple(V2(0.1234, 0.9876), V2(0.8765, 0.2345), 2.0),
tuple(V2(0.0000, 2.0000), V2(0.0000, 0.0000), 1.0),
tuple(V2(0.1234, 0.9876), V2(0.1234, 0.9876), 2.0),
tuple(V2(0.1234, 0.9876), V2(0.8765, 0.2345), 0.5),
tuple(V2(0.1234, 0.9876), V2(0.1234, 0.9876), 0.0)]) {
writefln("Through points:\n  %s   %s  and radius %f\n" ~
"You can construct the following circles:", t[]);
try {
writefln("  %s\n  %s\n",
} catch (ValueException v)
writefln("  ERROR: %s\n", v.msg);
}
}
```
Output:
```Through points:
immutable(V2)(0.1234, 0.9876)   immutable(V2)(0.8765, 0.2345)  and radius 2.000000
You can construct the following circles:
Circle(1.86311, 1.97421, 2)
Circle(-0.863212, -0.752112, 2)

Through points:
immutable(V2)(0, 2)   immutable(V2)(0, 0)  and radius 1.000000
You can construct the following circles:
Circle(0, 1, 1)
Circle(0, 1, 1)

Through points:
immutable(V2)(0.1234, 0.9876)   immutable(V2)(0.1234, 0.9876)  and radius 2.000000
You can construct the following circles:
ERROR: coincident points give infinite number of Circles

Through points:
immutable(V2)(0.1234, 0.9876)   immutable(V2)(0.8765, 0.2345)  and radius 0.500000
You can construct the following circles:
ERROR: separation of points > diameter

Through points:
immutable(V2)(0.1234, 0.9876)   immutable(V2)(0.1234, 0.9876)  and radius 0.000000
You can construct the following circles:
```

## Delphi

Library: System.Math
Translation of: C
```program Circles_of_given_radius_through_two_points;

{\$APPTYPE CONSOLE}

uses
System.SysUtils,
System.Types,
System.Math;

const
Cases: array[0..9] of TPointF = ((
x: 0.1234;
y: 0.9876
), (
x: 0.8765;
y: 0.2345
), (
x: 0.0000;
y: 2.0000
), (
x: 0.0000;
y: 0.0000
), (
x: 0.1234;
y: 0.9876
), (
x: 0.1234;
y: 0.9876
), (
x: 0.1234;
y: 0.9876
), (
x: 0.8765;
y: 0.2345
), (
x: 0.1234;
y: 0.9876
), (
x: 0.1234;
y: 0.9876
));
radii: array of double = [2.0, 1.0, 2.0, 0.5, 0.0];

procedure FindCircles(p1, p2: TPointF; radius: double);
var
separation, mirrorDistance: double;
begin
separation := p1.Distance(p2);
if separation = 0.0 then
begin
write(format(#10'No circles can be drawn through (%.4f,%.4f)', [p1.x, p1.y]))
else
write(format(#10'Infinitely many circles can be drawn through (%.4f,%.4f)',
[p1.x, p1.y]));
exit;
end;

if separation = 2 * radius then
begin
write(format(#10'Given points are opposite ends of a diameter of the circle with center (%.4f,%.4f) and radius %.4f',
[(p1.x + p2.x) / 2, (p1.y + p2.y) / 2, radius]));
exit;
end;

if separation > 2 * radius then
begin
write(format(#10'Given points are farther away from each other than a diameter of a circle with radius %.4f',
exit;
end;

mirrorDistance := sqrt(Power(radius, 2) - Power(separation / 2, 2));
write(#10'Two circles are possible.');
write(format(#10'Circle C1 with center (%.4f,%.4f), radius %.4f and Circle C2 with center (%.4f,%.4f), radius %.4f',
[(p1.x + p2.x) / 2 + mirrorDistance * (p1.y - p2.y) / separation, (p1.y + p2.y)
/ 2 + mirrorDistance * (p2.x - p1.x) / separation, radius, (p1.x + p2.x) / 2
- mirrorDistance * (p1.y - p2.y) / separation, (p1.y + p2.y) / 2 -
mirrorDistance * (p2.x - p1.x) / separation, radius]));

end;

begin
for var i := 0 to 4 do
begin
write(#10'Case ', i + 1,')');
findCircles(cases[2 * i], cases[2 * i + 1], radii[i]);
end;
end.
```

## EasyLang

Translation of: AWK
```func\$ fmt a b .
return "(" & a & " " & b & ")"
.
proc test m1x m1y m2x m2y r . .
print "Points: " & fmt m1x m1y & " " & fmt m2x m2y & " Radius: " & r
if r = 0
print "Radius of zero gives no circles"
print ""
return
.
x = (m2x - m1x) / 2
y = (m2y - m1y) / 2
bx = m1x + x
by = m1y + y
pb = sqrt (x * x + y * y)
if pb = 0
print "Coincident points give infinite circles"
print ""
return
.
if pb > r
print "Points are too far apart for the given radius"
print ""
return
.
cb = sqrt (r * r - pb * pb)
x1 = y * cb / pb
y1 = x * cb / pb
print "Circles: " & fmt (bx - x1) (by + y1) & " " & fmt (bx + x1) (by - y1)
print ""
.
test 0.1234 0.9876 0.8765 0.2345 2.0
test 0.0000 2.0000 0.0000 0.0000 1.0
test 0.1234 0.9876 0.1234 0.9876 2.0
test 0.1234 0.9876 0.8765 0.2345 0.5
test 0.1234 0.9876 0.1234 0.9876 0.0```

## Elixir

Translation of: Ruby
```defmodule RC do
def circle(p, p, r) when r>0.0 do
raise ArgumentError, message: "Infinite number of circles, points coincide."
end
def circle(p, p, r) when r==0.0 do
{px, py} = p
[{px, py, r}]
end
def circle({p1x,p1y}, {p2x,p2y}, r) do
{dx, dy} = {p2x-p1x, p2y-p1y}
q = :math.sqrt(dx*dx + dy*dy)
if q > 2*r do
raise ArgumentError, message: "Distance of points > diameter."
else
{x3, y3} = {(p1x+p2x) / 2, (p1y+p2y) / 2}
d = :math.sqrt(r*r - q*q/4)
Enum.uniq([{x3 - d*dy/q, y3 + d+dx/q, r}, {x3 + d*dy/q, y3 - d*dx/q, r}])
end
end
end

data = [{{0.1234, 0.9876}, {0.8765, 0.2345}, 2.0},
{{0.0000, 2.0000}, {0.0000, 0.0000}, 1.0},
{{0.1234, 0.9876}, {0.1234, 0.9876}, 2.0},
{{0.1234, 0.9876}, {0.8765, 0.2345}, 0.5},
{{0.1234, 0.9876}, {0.1234, 0.9876}, 0.0}]

Enum.each(data, fn {p1, p2, r} ->
IO.write "Given points:\n  #{inspect p1},\n  #{inspect p2}\n  and radius #{r}\n"
try do
circles = RC.circle(p1, p2, r)
IO.puts "You can construct the following circles:"
Enum.each(circles, fn circle -> IO.puts "  #{inspect circle}" end)
rescue
e in ArgumentError -> IO.inspect e
end
IO.puts ""
end)
```
Output:
```Given points:
{0.1234, 0.9876},
{0.8765, 0.2345}
You can construct the following circles:
{1.8631118016581893, 3.2459586888005014, 2.0}
{-0.8632118016581896, -0.7521118016581892, 2.0}

Given points:
{0.0, 2.0},
{0.0, 0.0}
You can construct the following circles:
{0.0, 1.0, 1.0}

Given points:
{0.1234, 0.9876},
{0.1234, 0.9876}
%ArgumentError{message: "Infinite number of circles, points coincide."}

Given points:
{0.1234, 0.9876},
{0.8765, 0.2345}
%ArgumentError{message: "Distance of points > diameter."}

Given points:
{0.1234, 0.9876},
{0.1234, 0.9876}
You can construct the following circles:
{0.1234, 0.9876, 0.0}
```

## ERRE

```PROGRAM CIRCLES

!
! for rosettacode.org
!

PROCEDURE CIRCLE_CENTER(X1,Y1,X2,Y2,R->MSG\$)
LOCAL D,W,X3,Y3

D=SQR((X2-X1)^2+(Y2-Y1)^2)
IF D=0 THEN
MSG\$="NO CIRCLES CAN BE DRAWN, POINTS ARE IDENTICAL"
EXIT PROCEDURE
END IF
X3=(X1+X2)/2  Y3=(Y1+Y2)/2

W=R^2-(D/2)^2
IF W<0 THEN
MSG\$="NO SOLUTION"
EXIT PROCEDURE
END IF
CX1=X3+SQR(W)*(Y1-Y2)/D   CY1=Y3+SQR(W)*(X2-X1)/D
CX2=X3-SQR(W)*(Y1-Y2)/D   CY2=Y3-SQR(W)*(X2-X1)/D
IF D=R*2 THEN
MSG\$="POINTS ARE OPPOSITE ENDS OF A DIAMETER CENTER = "+STR\$(CX1)+","+STR\$(CY1)
EXIT PROCEDURE
END IF
IF D>R*2 THEN
MSG\$="POINTS ARE TOO FAR"
EXIT PROCEDURE
END IF
IF R<=0 THEN
EXIT PROCEDURE
END IF
MSG\$=STR\$(CX1)+","+STR\$(CY1)+" & "+STR\$(CX2)+","+STR\$(CY2)
END PROCEDURE

BEGIN
DATA(0.1234,0.9876,0.8765,0.2345,2.0)
DATA(0.0000,2.0000,0.0000,0.0000,1.0)
DATA(0.1234,0.9876,0.1234,0.9876,2.0)
DATA(0.1234,0.9876,0.8765,0.2345,0.5)
DATA(0.1234,0.9876,0.1234,0.9876,0.0)

FOR I%=1 TO 5 DO
PRINT(MSG\$)
END FOR
END PROGRAM
```

## F#

```open System

let add (a:double, b:double) (x:double, y:double) = (a + x, b + y)
let sub (a:double, b:double) (x:double, y:double) = (a - x, b - y)
let magSqr (a:double, b:double) = a * a + b * b
let mag a:double = Math.Sqrt(magSqr a)
let mul (a:double, b:double) c = (a * c, b * c)
let div2 (a:double, b:double) c = (a / c, b / c)
let perp (a:double, b:double) = (-b, a)
let norm a = div2 a (mag a)

let circlePoints p q (radius:double) =
let diameter = radius * 2.0
let pq = sub p q
let magPQ = mag pq
let midpoint = div2 (add p q) 2.0
let halfPQ = magPQ / 2.0
let midC = mul (norm (perp pq)) magMidC
let center1 = add midpoint midC
let center2 = sub midpoint midC
if radius = 0.0 then None
else if p = q then None
else if diameter < magPQ then None
else Some (center1, center2)

[<EntryPoint>]
let main _ =
printfn "%A" (circlePoints (0.1234, 0.9876) (0.8765, 0.2345) 2.0)
printfn "%A" (circlePoints (0.0, 2.0) (0.0, 0.0) 1.0)
printfn "%A" (circlePoints (0.1234, 0.9876) (0.1234, 0.9876) 2.0)
printfn "%A" (circlePoints (0.1234, 0.9876) (0.8765, 0.2345) 0.5)
printfn "%A" (circlePoints (0.1234, 0.9876) (0.1234, 0.1234) 0.0)

0 // return an integer exit code
```
Output:
```Some ((-0.8632118017, -0.7521118017), (1.863111802, 1.974211802))
Some ((0.0, 1.0), (0.0, 1.0))
<null>
<null>
<null>```

## Factor

```USING: accessors combinators combinators.short-circuit
formatting io kernel literals locals math math.distances
math.functions prettyprint sequences strings ;
IN: rosetta-code.circles
DEFER: find-circles

! === Input ====================================================

TUPLE: input p1 p2 r ;

CONSTANT: test-cases {
T{ input f { 0.1234 0.9876 } { 0.8765 0.2345 } 2 }
T{ input f { 0 2 } { 0 0 } 1 }
T{ input f { 0.1234 0.9876 } { 0.1234 0.9876 } 2 }
T{ input f { 0.1234 0.9876 } { 0.8765 0.2345 } 0.5 }
T{ input f { 0.1234 0.9876 } { 0.1234 0.9876 } 0 }
}

! === Edge case handling =======================================

CONSTANT: infinite
"there could be an infinite number of circles."

CONSTANT: too-far
"points are too far apart to form circles."

: coincident? ( input -- ? ) [ p1>> ] [ p2>> ] bi = ;

: degenerate? ( input -- ? )
{ [ r>> zero? ] [ coincident? ] } 1&& ;

: infinite? ( input -- ? )
{ [ r>> zero? not ] [ coincident? ] } 1&& ;

: too-far? ( input -- ? )
[ r>> 2 * ] [ p1>> ] [ p2>> ] tri euclidian-distance < ;

: degenerate ( input -- str )
p1>> [ first ] [ second ] bi
"one degenerate circle found at (%.4f, %.4f).\n" sprintf ;

: check-input ( input -- obj )
{
{ [ dup infinite?   ] [ drop infinite ] }
{ [ dup too-far?    ] [ drop too-far  ] }
{ [ dup degenerate? ] [ degenerate    ] }
[ find-circles ]
} cond ;

! === Program Logic ============================================

:: (circle-coords) ( a b c r q quot -- x )
a r sq q 2 / sq - sqrt b c - * q / quot call ; inline

: circle-coords ( quot -- x y )
[ + ] over [ - ] [ [ call ] dip (circle-coords) ] 2bi@ ;
inline

:: find-circles ( input -- circles )
input [ r>> ] [ p1>> ] [ p2>> ] tri    :> ( r p1 p2 )
p1 p2 [ [ first ] [ second ] bi ] bi@  :> ( x1 y1 x2 y2 )
x1 x2 y1 y2 [ + 2 / ] 2bi@             :> ( x3 y3 )
p1 p2 euclidian-distance               :> q
[ x3 y1 y2 r q ]
[ y3 x2 x1 r q ] [ circle-coords ] bi@ :> ( x w y z )
{ x y } { w z } = { { x y } } { { w z } { x y } } ? ;

! === Output ===================================================

: .point ( seq -- )
[ first ] [ second ] bi "(%.4f, %.4f)" printf ;

: .given ( input -- )
[ r>> ] [ p2>> ] [ p1>> ] tri
"Given points " write .point ", " write .point
", and radius %.2f,\n" printf ;

: .one ( seq -- )
first "one circle found at " write .point "." print ;

: .two ( seq -- )
[ first ] [ second ] bi "two circles found at " write
.point " and " write .point "." print ;

: .circles ( seq -- ) dup length 1 = [ .one ] [ .two ] if ;

! === Main word ================================================

: circles-demo ( -- )
test-cases [
dup .given check-input dup string?
[ print ] [ .circles ] if nl
] each ;

MAIN: circles-demo
```
Output:
```Given points (0.1234, 0.9876), (0.8765, 0.2345), and radius 2.00,
two circles found at (1.8631, 1.9742) and (-0.8632, -0.7521).

Given points (0.0000, 2.0000), (0.0000, 0.0000), and radius 1.00,
one circle found at (0.0000, 1.0000).

Given points (0.1234, 0.9876), (0.1234, 0.9876), and radius 2.00,
there could be an infinite number of circles.

Given points (0.1234, 0.9876), (0.8765, 0.2345), and radius 0.50,
points are too far apart to form circles.

Given points (0.1234, 0.9876), (0.1234, 0.9876), and radius 0.00,
one degenerate circle found at (0.1234, 0.9876).
```

## Fortran

```! Implemented by Anant Dixit (Nov. 2014)
! Transpose elements in find_center to obtain correct results. R.N. McLean (Dec 2017)
program circles
implicit none
double precision :: P1(2), P2(2), R

P1 = (/0.1234d0, 0.9876d0/)
P2 = (/0.8765d0,0.2345d0/)
R = 2.0d0
call print_centers(P1,P2,R)

P1 = (/0.0d0, 2.0d0/)
P2 = (/0.0d0,0.0d0/)
R = 1.0d0
call print_centers(P1,P2,R)

P1 = (/0.1234d0, 0.9876d0/)
P2 = (/0.1234d0, 0.9876d0/)
R = 2.0d0
call print_centers(P1,P2,R)

P1 = (/0.1234d0, 0.9876d0/)
P2 = (/0.8765d0, 0.2345d0/)
R = 0.5d0
call print_centers(P1,P2,R)

P1 = (/0.1234d0, 0.9876d0/)
P2 = (/0.1234d0, 0.9876d0/)
R = 0.0d0
call print_centers(P1,P2,R)
end program circles

subroutine print_centers(P1,P2,R)
implicit none
double precision :: P1(2), P2(2), R, Center(2,2)
integer :: Res
call test_inputs(P1,P2,R,Res)
write(*,*)
write(*,'(A10,F7.4,A1,F7.4)') 'Point1  : ', P1(1), ' ', P1(2)
write(*,'(A10,F7.4,A1,F7.4)') 'Point2  : ', P2(1), ' ', P2(2)
if(Res.eq.1) then
write(*,*) 'Same point because P1=P2 and r=0.'
elseif(Res.eq.2) then
write(*,*) 'No circles can be drawn because r=0.'
elseif(Res.eq.3) then
write(*,*) 'Infinite circles because P1=P2 for non-zero radius.'
elseif(Res.eq.4) then
write(*,*) 'No circles with given r can be drawn because points are far apart.'
elseif(Res.eq.0) then
call find_center(P1,P2,R,Center)
if(Center(1,1).eq.Center(2,1) .and. Center(1,2).eq.Center(2,2)) then
write(*,*) 'Points lie on the diameter. A single circle can be drawn.'
write(*,'(A10,F7.4,A1,F7.4)') 'Center  : ', Center(1,1), ' ', Center(1,2)
else
write(*,*) 'Two distinct circles found.'
write(*,'(A10,F7.4,A1,F7.4)') 'Center1 : ', Center(1,1), ' ', Center(1,2)
write(*,'(A10,F7.4,A1,F7.4)') 'Center2 : ', Center(2,1), ' ', Center(2,2)
end if
elseif(Res.lt.0) then
write(*,*) 'Incorrect value for r.'
end if
write(*,*)
end subroutine print_centers

subroutine test_inputs(P1,P2,R,Res)
implicit none
double precision :: P1(2), P2(2), R, dist
integer :: Res
if(R.lt.0.0d0) then
Res = -1
return
elseif(R.eq.0.0d0 .and. P1(1).eq.P2(1) .and. P1(2).eq.P2(2)) then
Res = 1
return
elseif(R.eq.0.0d0) then
Res = 2
return
elseif(P1(1).eq.P2(1) .and. P1(2).eq.P2(2)) then
Res = 3
return
else
dist = sqrt( (P1(1)-P2(1))**2 + (P1(2)-P2(2))**2 )
if(dist.gt.2.0d0*R) then
Res = 4
return
else
Res = 0
return
end if
end if
end subroutine test_inputs

subroutine find_center(P1,P2,R,Center)
implicit none
double precision :: P1(2), P2(2), MP(2), Center(2,2), R, dm, dd
MP = (P1 + P2)/2.0d0
dm = sqrt((P1(1) - P2(1))**2 + (P1(2) - P2(2))**2)
dd = sqrt(R**2 - (dm/2.0d0)**2)
Center(1,1) = MP(1) - dd*(P2(2) - P1(2))/dm
Center(1,2) = MP(2) + dd*(P2(1) - P1(1))/dm

Center(2,1) = MP(1) + dd*(P2(2) - P1(2))/dm
Center(2,2) = MP(2) - dd*(P2(1) - P1(1))/dm
end subroutine find_center
```
Output:
```
Point1  :  0.1234  0.9876
Point2  :  0.8765  0.2345
Two distinct circles found.
Center1 :  1.8631  1.9742
Center2 : -0.8632 -0.7521

Point1  :  0.0000  2.0000
Point2  :  0.0000  0.0000
Points lie on the diameter. A single circle can be drawn.
Center  :  0.0000  1.0000

Point1  :  0.1234  0.9876
Point2  :  0.1234  0.9876
Infinite circles because P1=P2 for non-zero radius.

Point1  :  0.1234  0.9876
Point2  :  0.8765  0.2345
No circles with given r can be drawn because points are far apart.

Point1  :  0.1234  0.9876
Point2  :  0.1234  0.9876
Same point because P1=P2 and r=0.
```

### Using complex numbers

Fortran 66 made standard the availability of complex number arithmetic. This version however takes advantage of facilities offered in F90 so as to perform some array-based arithmetic, though the opportunities in this small routine are thin: two statements become one (look for CMPLX). More seriously, the MODULE facility allows the definition of an array SQUAWK which contains an explanatory text associated with each return code. The routine has a troublesome variety of possible odd conditions to report. An older approach would be to have a return message CHARACTER variable to present the remark, at the cost of filling up that variable with text every time. By returning an integer code, less effort is required, but there is no explication of the return codes. One could still have an array of messages (and prior to F90, array index counting started at one only, so no starting with -3 for errorish codes) but making that array available would require some sort of COMMON storage. The MODULE facility eases this problem.

```      MODULE GEOMETRY	!Limited scope.
CHARACTER*(*) SQUAWK(-3:2)	!Holds a schedule of complaints.
PARAMETER (SQUAWK = (/		!According to what might go wrong.
3  "No circles: points are more than 2R apart.",
2  "Innumerable circles: co-incident points, R > 0.",
1  "One 'circle', centred on the co-incident points. R is zero!",
o  "No circles! R is negative!",
1  "One circle: points are 2R apart.",
2  "Two circles."/))		!This last is the hoped-for state.
CONTAINS	!Now for the action.
SUBROUTINE BUBBLE(P,R,N)	!Finds circles of radius R passing through two points.
COMPLEX P(2)	!The two points. Results returned here.
INTEGER N	!Indicates how many centres are valid.
COMPLEX MID,DP	!Geometrical assistants.
DP = (P(2) - P(1))/2	!Or, the other way around.
D = ABS(DP)		!Half the separation is useful.
IF (R.LT.0) THEN	!Is the specified radius silly?
N =  0			!Yes. No circles, then.
ELSE IF (D.EQ.0) THEN	!Any distance between the points?
IF (R.EQ.0) THEN		!No. Zero radius?
N = -1				!Yes. So a degenerate "circle" of zero radius.
ELSE			!A negative radius being tested for above,
N = -2				!A swirl of circles around the midpoint.
END IF		!So much for co-incident points.
ELSE IF (D.GT.R) THEN	!Points too far apart?
N = -3			!A circle of radius R can't reach them.
ELSE IF (D.EQ.R) THEN	!Maximum separation for R?
N = 1			!Yes. The two circles lie atop each other.
P(1) = (P(1) + P(2))/2	!Both centres are on the midpoint, but N = 1.
ELSE			!Finally, the ordinary case.
N = 2			!Two circles.
MID = (P(1) + P(2))/2	!Midway between the two points.
D = SQRT((R/D)**2 - 1)	!Rescale vector DP.
P = MID + DP*CMPLX(0,(/+D,-D/))	!Array (0,+D), (0,-D)
END IF				!P(1) = DP*CMPLX(0,+D) and P(2) = DP*CMPLX(0,-D)
END SUBROUTINE BUBBLE	!Careful! P and N are modified.
END MODULE GEOMETRY	!Not much.

PROGRAM POKE	!A tester.
USE GEOMETRY	!Useful to I. Newton.
COMPLEX P(2)		!A pair of points.
REAL PP(4)		!Also a pair.
EQUIVALENCE (P,PP)	!Since free-format input likes (x,y), not x,y
REAL R			!This is not complex.
INTEGER MSG,IN		!I/O unit numbers.
MSG = 6			!Standard output.
OPEN (MSG, RECL = 120)	!For "formatted" files, this length is in characters.
IN = 10			!For the disc file holding the test data.
WRITE (MSG,1)		!Announce.
1 FORMAT ("Given two points and a radius, find the centres "
1 "of circles of that radius passing through those points.")

OPEN (IN,FILE="Circle.csv", STATUS = "OLD", ACTION="READ")	!Have data, will compute.
10 READ (IN,*,END = 20) PP,R		!Get two points and a radius.
WRITE (MSG,*)			!Set off.
WRITE (MSG,*) P,R			!Show the input.
CALL BUBBLE(P,R,N)		!Calculate.
WRITE (MSG,*) P(1:N),SQUAWK(N)	!Show results.
GO TO 10				!Try it again.

20 CLOSE(IN)		!Finihed with input.
END	!Finished.
```

Results: little attempt has been made to present a fancy layout, "free-format" output does well enough. Notably, complex numbers are presented in brackets with a comma as (x,y); a FORMAT statement version would have to supply those decorations. Free-format input also expects such bracketing when reading complex numbers. The supplied data format however does not include the brackets and so is improper. Suitable data would be

```(0.1234, 0.9876)    (0.8765, 0.2345)    2.0
(0.0000, 2.0000)    (0.0000, 0.0000)    1.0
(0.1234, 0.9876)    (0.1234, 0.9876)    2.0
(0.1234, 0.9876)    (0.8765, 0.2345)    0.5
(0.1234, 0.9876)    (0.1234, 0.9876)    0.0
```

The free-format input style allows spaces, a comma (with or without spaces), and even a tab as delimiters between data, but does not allow implicit delimiters so a sequence such as 2017-12-29 (a standard date format) would be rejected. Because the style of the supplied data does not include the brackets, when complex numbers are read from such an input stream, they are taken to be real numbers only so that each real number is deemed a complex number of the form (x,0); in this case the second number would be taken as being the real part of the second complex number. A mess results.

By using the EQUIVALENCE statement, array PP can be read via the free-format protocol, and so the first four numbers will be placed in array PP, which just happens to be the same storage area as the array P of complex numbers. This of course means that should proper bracketed complex numbers be presented as input, a different mess results.

Output:

```Given two points and a radius, find the centres of circles of that radius passing through those points.

(0.1234000,0.9876000) (0.8765000,0.2345000)   2.000000
(1.863112,1.974212) (-0.8632119,-0.7521119) Two circles.

(0.0000000E+00,2.000000) (0.0000000E+00,0.0000000E+00)   1.000000
(0.0000000E+00,1.000000) One circle: points are 2R apart.

(0.1234000,0.9876000) (0.1234000,0.9876000)   2.000000
Innumerable circles: co-incident points, R > 0.

(0.1234000,0.9876000) (0.8765000,0.2345000)  0.5000000
No circles: points are more than 2R apart.

(0.1234000,0.9876000) (0.1234000,0.9876000)  0.0000000E+00
One 'circle', centred on the co-incident points. R is zero!
```

## FreeBASIC

```Type Point
As Double x,y
Declare Property length As Double
End Type

Property point.length As Double
Return Sqr(x*x+y*y)
End Property

Sub circles(p1 As Point,p2 As Point,radius As Double)
Var ctr=Type<Point>((p1.x+p2.x)/2,(p1.y+p2.y)/2)
Var half=Type<Point>(p1.x-ctr.x,p1.y-ctr.y)
Var lenhalf=half.length
If radius<lenhalf Then Print "Can't solve":Print:Exit Sub
If lenhalf=0 Then Print "Points are the same":Print:Exit Sub
Var rot= Type<Point>(-dist*(p1.y-ctr.y) +ctr.x,dist*(p1.x-ctr.x) +ctr.y)
Print " -> Circle 1 ("&rot.x;","&rot.y;")"
rot= Type<Point>(-(rot.x-ctr.x) +ctr.x,-((rot.y-ctr.y)) +ctr.y)
Print" -> Circle 2 ("&rot.x;","&rot.y;")"
Print
End Sub

Dim As Point p1=(.1234,.9876),p2=(.8765,.2345)
circles(p1,p2,2)
p1=Type<Point>(0,2):p2=Type<Point>(0,0)
circles(p1,p2,1)
p1=Type<Point>(.1234,.9876):p2=p1
circles(p1,p2,2)
p1=Type<Point>(.1234,.9876):p2=Type<Point>(.8765,.2345)
circles(p1,p2,.5)
p1=Type<Point>(.1234,.9876):p2=p1
circles(p1,p2,0)

Sleep```
Output:
```Points (0.1234,0.9876),(0.8765,0.2345), Rad  2
-> Circle 1 (-0.8632118016581893,-0.7521118016581889)
-> Circle 2 (1.863111801658189,1.974211801658189)

-> Circle 1 (0,1)
-> Circle 2 (0,1)

Points are the same

Can't solve

Points are the same```

## Go

```package main

import (
"fmt"
"math"
)

var (
Two  = "Two circles."
R0   = "R==0.0 does not describe circles."
Co   = "Coincident points describe an infinite number of circles."
CoR0 = "Coincident points with r==0.0 describe a degenerate circle."
Diam = "Points form a diameter and describe only a single circle."
Far  = "Points too far apart to form circles."
)

type point struct{ x, y float64 }

func circles(p1, p2 point, r float64) (c1, c2 point, Case string) {
if p1 == p2 {
if r == 0 {
return p1, p1, CoR0
}
Case = Co
return
}
if r == 0 {
return p1, p2, R0
}
dx := p2.x - p1.x
dy := p2.y - p1.y
q := math.Hypot(dx, dy)
if q > 2*r {
Case = Far
return
}
m := point{(p1.x + p2.x) / 2, (p1.y + p2.y) / 2}
if q == 2*r {
return m, m, Diam
}
d := math.Sqrt(r*r - q*q/4)
ox := d * dx / q
oy := d * dy / q
return point{m.x - oy, m.y + ox}, point{m.x + oy, m.y - ox}, Two
}

var td = []struct {
p1, p2 point
r      float64
}{
{point{0.1234, 0.9876}, point{0.8765, 0.2345}, 2.0},
{point{0.0000, 2.0000}, point{0.0000, 0.0000}, 1.0},
{point{0.1234, 0.9876}, point{0.1234, 0.9876}, 2.0},
{point{0.1234, 0.9876}, point{0.8765, 0.2345}, 0.5},
{point{0.1234, 0.9876}, point{0.1234, 0.9876}, 0.0},
}

func main() {
for _, tc := range td {
fmt.Println("p1: ", tc.p1)
fmt.Println("p2: ", tc.p2)
fmt.Println("r: ", tc.r)
c1, c2, Case := circles(tc.p1, tc.p2, tc.r)
fmt.Println("  ", Case)
switch Case {
case CoR0, Diam:
fmt.Println("   Center: ", c1)
case Two:
fmt.Println("   Center 1: ", c1)
fmt.Println("   Center 2: ", c2)
}
fmt.Println()
}
}
```
Output:
```p1:  {0.1234 0.9876}
p2:  {0.8765 0.2345}
r:  2
Two circles.
Center 1:  {1.8631118016581891 1.974211801658189}
Center 2:  {-0.8632118016581893 -0.752111801658189}

p1:  {0 2}
p2:  {0 0}
r:  1
Points form a diameter and describe only a single circle.
Center:  {0 1}

p1:  {0.1234 0.9876}
p2:  {0.1234 0.9876}
r:  2
Coincident points describe an infinite number of circles.

p1:  {0.1234 0.9876}
p2:  {0.8765 0.2345}
r:  0.5
Points too far apart to form circles.

p1:  {0.1234 0.9876}
p2:  {0.1234 0.9876}
r:  0
Coincident points with r==0.0 describe a degenerate circle.
Center:  {0.1234 0.9876}
```

## Groovy

Translation of: Java
```class Circles {
private static class Point {
private final double x, y

Point(Double x, Double y) {
this.x = x
this.y = y
}

double distanceFrom(Point other) {
double dx = x - other.x
double dy = y - other.y
return Math.sqrt(dx * dx + dy * dy)
}

@Override
boolean equals(Object other) {
//if (this == other) return true
if (other == null || getClass() != other.getClass()) return false
Point point = (Point) other
return x == point.x && y == point.y
}

@Override
String toString() {
return String.format("(%.4f, %.4f)", x, y)
}
}

private static Point[] findCircles(Point p1, Point p2, double r) {
if (r < 0.0) throw new IllegalArgumentException("the radius can't be negative")
if (r == 0.0.toDouble() && p1 != p2) throw new IllegalArgumentException("no circles can ever be drawn")
if (r == 0.0.toDouble()) return [p1, p1]
if (Objects.equals(p1, p2)) throw new IllegalArgumentException("an infinite number of circles can be drawn")
double distance = p1.distanceFrom(p2)
double diameter = 2.0 * r
if (distance > diameter) throw new IllegalArgumentException("the points are too far apart to draw a circle")
Point center = new Point((p1.x + p2.x) / 2.0, (p1.y + p2.y) / 2.0)
if (distance == diameter) return [center, center]
double mirrorDistance = Math.sqrt(r * r - distance * distance / 4.0)
double dx = (p2.x - p1.x) * mirrorDistance / distance
double dy = (p2.y - p1.y) * mirrorDistance / distance
return [
new Point(center.x - dy, center.y + dx),
new Point(center.x + dy, center.y - dx)
]
}

static void main(String[] args) {
Point[] p = [
new Point(0.1234, 0.9876),
new Point(0.8765, 0.2345),
new Point(0.0000, 2.0000),
new Point(0.0000, 0.0000)
]
Point[][] points = [
[p[0], p[1]],
[p[2], p[3]],
[p[0], p[0]],
[p[0], p[1]],
[p[0], p[0]],
]
double[] radii = [2.0, 1.0, 2.0, 0.5, 0.0]
for (int i = 0; i < radii.length; ++i) {
Point p1 = points[i][0]
Point p2 = points[i][1]
printf("For points %s and %s with radius %f\n", p1, p2, r)
try {
Point[] circles = findCircles(p1, p2, r)
Point c1 = circles[0]
Point c2 = circles[1]
if (Objects.equals(c1, c2)) {
printf("there is just one circle with center at %s\n", c1)
} else {
printf("there are two circles with centers at %s and %s\n", c1, c2)
}
} catch (IllegalArgumentException ex) {
println(ex.getMessage())
}
println()
}
}
}
```
Output:
```For points (0.1234, 0.9876) and (0.8765, 0.2345) with radius 2.000000
there are two circles with centers at (1.8631, 1.9742) and (-0.8632, -0.7521)

For points (0.0000, 2.0000) and (0.0000, 0.0000) with radius 1.000000
there is just one circle with center at (0.0000, 1.0000)

For points (0.1234, 0.9876) and (0.1234, 0.9876) with radius 2.000000
an infinite number of circles can be drawn

For points (0.1234, 0.9876) and (0.8765, 0.2345) with radius 0.500000
the points are too far apart to draw a circle

For points (0.1234, 0.9876) and (0.1234, 0.9876) with radius 0.000000
there is just one circle with center at (0.1234, 0.9876)```

```add (a, b) (x, y) = (a + x, b + y)
sub (a, b) (x, y) = (a - x, b - y)
magSqr (a, b)     = (a ^^ 2) + (b ^^ 2)
mag a             = sqrt \$ magSqr a
mul (a, b) c      = (a * c, b * c)
div2 (a, b) c     = (a / c, b / c)
perp (a, b)       = (negate b, a)
norm a            = a `div2` mag a

circlePoints :: (Ord a, Floating a) =>
(a, a) -> (a, a) -> a -> Maybe ((a, a), (a, a))
| radius == 0      = Nothing
| p == q           = Nothing
| diameter < magPQ = Nothing
| otherwise        = Just (center1, center2)
where
pq       = p `sub` q
magPQ    = mag pq
midpoint = (p `add` q) `div2` 2
halfPQ   = magPQ / 2
magMidC  = sqrt . abs \$ (radius ^^ 2) - (halfPQ ^^ 2)
midC     = (norm \$ perp pq) `mul` magMidC
center2  = midpoint `sub` midC

uncurry3 f (a, b, c) = f a b c

main :: IO ()
main = mapM_ (print . uncurry3 circlePoints)
[((0.1234, 0.9876), (0.8765, 0.2345), 2),
((0     , 2     ), (0     , 0     ), 1),
((0.1234, 0.9876), (0.1234, 0.9876), 2),
((0.1234, 0.9876), (0.8765, 0.2345), 0.5),
((0.1234, 0.9876), (0.1234, 0.1234), 0)]
```
Output:
```Just ((-0.8632118016581896,-0.7521118016581892),(1.8631118016581893,1.974211801658189))
Just ((0.0,1.0),(0.0,1.0))
Nothing
Nothing
Nothing```

## Icon and Unicon

Translation of: AutoHotKey

Works in both languages.

```procedure main()
A := [ [0.1234, 0.9876,   0.8765, 0.2345,   2.0],
[0.0000, 2.0000,   0.0000, 0.0000,   1.0],
[0.1234, 0.9876,   0.1234, 0.9876,   2.0],
[0.1234, 0.9876,   0.9765, 0.2345,   0.5],
[0.1234, 0.9876,   0.1234, 0.9876,   0.0] ]
every write(cCenter!!A)
end

procedure cCenter(x1,y1, x2,y2, r)
if r <= 0 then return "Illegal radius"
r2 := r*2
d := ((x2-x1)^2 + (y2-y1)^2)^0.5
if d = 0 then return "Identical points, infinite number of circles"
if d > r2 then return "No circles possible"
z   := (r^2-(d/2.0)^2)^0.5
x3  := (x1+x2)/2.0;     y3 := (y1+y2)/2.0
cx1 := x3+z*(y1-y2)/d; cy1 := y3+z*(x2-x1)/d
cx2 := x3-z*(y1-y2)/d; cy2 := y3-z*(x2-x1)/d
if d = r2 then return "Single circle at ("||cx1||","||cy1||")"
return "("||cx1||","||cy1||") and ("||cx2||","||cy2||")"
end
```
Output:
```->cgr
(1.863111801658189,1.974211801658189) and (-0.8632118016581896,-0.7521118016581892)
Single circle at (0.0,1.0)
Identical points, infinite number of circles
No circles possible
->
```

## J

2D computations are often easier using the complex plane.

```average =: +/ % #

circles =: verb define"1
'P0 P1 R' =. (j./"1)_2[\y NB. Use complex plane
C =. P0 average@:, P1
SEPARATION =. P0 |@- P1
if. 0 = SEPARATION do.
if. 0 = R do. 'Degenerate point at ' , BAD
else. 'Any center at a distance ' , (":R) , ' from ' , BAD , ' works.'
end.
elseif. SEPARATION (> +:) R do. 'No solutions.'
elseif. SEPARATION (= +:) R do. 'Duplicate solutions with center at ' , BAD
elseif. 1 do.
ORTHOGONAL_DISTANCE =. R * 1 o. _2 o. R %~ | C - P0
UNIT =: P1 *@:- P0
OFFSETS =: ORTHOGONAL_DISTANCE * UNIT * j. _1 1
C +.@:+ OFFSETS
end.
)

INPUT=: ".;._2]0 :0
0.1234 0.9876 0.8765 0.2345   2
0      2      0      0   1
0.1234 0.9876 0.1234 0.9876   2
0.1234 0.9876 0.8765 0.2345 0.5
0.1234 0.9876 0.1234 0.9876   0
)

('x0 y0 x1 y1 r' ; 'center'),(;circles)"1 INPUT
┌───────────────────────────────┬────────────────────────────────────────────────────┐
│x0 y0 x1 y1 r                  │center                                              │
├───────────────────────────────┼────────────────────────────────────────────────────┤
│0.1234 0.9876 0.8765 0.2345 2  │_0.863212 _0.752112                                 │
│                               │  1.86311   1.97421                                 │
├───────────────────────────────┼────────────────────────────────────────────────────┤
│0 2 0 0 1                      │Duplicate solutions with center at 0 1              │
├───────────────────────────────┼────────────────────────────────────────────────────┤
│0.1234 0.9876 0.1234 0.9876 2  │Any center at a distance 2 from 0.1234 0.9876 works.│
├───────────────────────────────┼────────────────────────────────────────────────────┤
│0.1234 0.9876 0.8765 0.2345 0.5│No solutions.                                       │
├───────────────────────────────┼────────────────────────────────────────────────────┤
│0.1234 0.9876 0.1234 0.9876 0  │Degenerate point at 0.1234 0.9876                   │
└───────────────────────────────┴────────────────────────────────────────────────────┘
```

## Java

Translation of: Kotlin
```import java.util.Objects;

public class Circles {
private static class Point {
private final double x, y;

public Point(Double x, Double y) {
this.x = x;
this.y = y;
}

public double distanceFrom(Point other) {
double dx = x - other.x;
double dy = y - other.y;
return Math.sqrt(dx * dx + dy * dy);
}

@Override
public boolean equals(Object other) {
if (this == other) return true;
if (other == null || getClass() != other.getClass()) return false;
Point point = (Point) other;
return x == point.x && y == point.y;
}

@Override
public String toString() {
return String.format("(%.4f, %.4f)", x, y);
}
}

private static Point[] findCircles(Point p1, Point p2, double r) {
if (r < 0.0) throw new IllegalArgumentException("the radius can't be negative");
if (r == 0.0 && p1 != p2) throw new IllegalArgumentException("no circles can ever be drawn");
if (r == 0.0) return new Point[]{p1, p1};
if (Objects.equals(p1, p2)) throw new IllegalArgumentException("an infinite number of circles can be drawn");
double distance = p1.distanceFrom(p2);
double diameter = 2.0 * r;
if (distance > diameter) throw new IllegalArgumentException("the points are too far apart to draw a circle");
Point center = new Point((p1.x + p2.x) / 2.0, (p1.y + p2.y) / 2.0);
if (distance == diameter) return new Point[]{center, center};
double mirrorDistance = Math.sqrt(r * r - distance * distance / 4.0);
double dx = (p2.x - p1.x) * mirrorDistance / distance;
double dy = (p2.y - p1.y) * mirrorDistance / distance;
return new Point[]{
new Point(center.x - dy, center.y + dx),
new Point(center.x + dy, center.y - dx)
};
}

public static void main(String[] args) {
Point[] p = new Point[]{
new Point(0.1234, 0.9876),
new Point(0.8765, 0.2345),
new Point(0.0000, 2.0000),
new Point(0.0000, 0.0000)
};
Point[][] points = new Point[][]{
{p[0], p[1]},
{p[2], p[3]},
{p[0], p[0]},
{p[0], p[1]},
{p[0], p[0]},
};
double[] radii = new double[]{2.0, 1.0, 2.0, 0.5, 0.0};
for (int i = 0; i < radii.length; ++i) {
Point p1 = points[i][0];
Point p2 = points[i][1];
System.out.printf("For points %s and %s with radius %f\n", p1, p2, r);
try {
Point[] circles = findCircles(p1, p2, r);
Point c1 = circles[0];
Point c2 = circles[1];
if (Objects.equals(c1, c2)) {
System.out.printf("there is just one circle with center at %s\n", c1);
} else {
System.out.printf("there are two circles with centers at %s and %s\n", c1, c2);
}
} catch (IllegalArgumentException ex) {
System.out.println(ex.getMessage());
}
System.out.println();
}
}
}
```
Output:
```For points (0.1234, 0.9876) and (0.8765, 0.2345) with radius 2.000000
there are two circles with centers at (1.8631, 1.9742) and (-0.8632, -0.7521)

For points (0.0000, 2.0000) and (0.0000, 0.0000) with radius 1.000000
there is just one circle with center at (0.0000, 1.0000)

For points (0.1234, 0.9876) and (0.1234, 0.9876) with radius 2.000000
an infinite number of circles can be drawn

For points (0.1234, 0.9876) and (0.8765, 0.2345) with radius 0.500000
the points are too far apart to draw a circle

For points (0.1234, 0.9876) and (0.1234, 0.9876) with radius 0.000000
there is just one circle with center at (0.1234, 0.9876)```

## JavaScript

#### ES6

```const hDist = (p1, p2) => Math.hypot(...p1.map((e, i) => e - p2[i])) / 2;
const pAng = (p1, p2) => Math.atan(p1.map((e, i) => e - p2[i]).reduce((p, c) => c / p, 1));
const solveF = (p, r) => t => [r*Math.cos(t) + p[0], r*Math.sin(t) + p[1]];
const diamPoints = (p1, p2) => p1.map((e, i) => e + (p2[i] - e) / 2);

const findC = (...args) => {
const [p1, p2, s] = args;
const solve = solveF(p1, s);
const halfDist = hDist(p1, p2);

let msg = `p1: \${p1}, p2: \${p2}, r:\${s} Result: `;
switch (Math.sign(s - halfDist)) {
case 0:
msg += s ? `Points on diameter. Circle at: \${diamPoints(p1, p2)}` :
break;
case 1:
if (!halfDist) {
msg += 'Coincident point. Infinite solutions';
}
else {
let theta = pAng(p1, p2);
let theta2 = Math.acos(halfDist / s);
[1, -1].map(e => solve(theta + e * theta2)).forEach(
e => msg += `Circle at \${e} `);
}
break;
case -1:
msg += 'No intersection. Points further apart than circle diameter';
break;
}
return msg;
};

[
[[0.1234, 0.9876], [0.8765, 0.2345], 2.0],
[[0.0000, 2.0000], [0.0000, 0.0000], 1.0],
[[0.1234, 0.9876], [0.1234, 0.9876], 2.0],
[[0.1234, 0.9876], [0.8765, 0.2345], 0.5],
[[0.1234, 0.9876], [0.1234, 0.9876], 0.0]
].forEach((t,i) => console.log(`Test: \${i}: \${findC(...t)}`));
```

Output:

```Test: 0: p1: 0.1234,0.9876, p2: 0.8765,0.2345, r:2 Result: Circle at 1.8631118016581891,1.974211801658189 Circle at -0.863211801658189,-0.7521118016581889
Test: 1: p1: 0,2, p2: 0,0, r:1 Result: Points on diameter. Circle at: 0,1
Test: 2: p1: 0.1234,0.9876, p2: 0.1234,0.9876, r:2 Result: Coincident point. Infinite solutions
Test: 3: p1: 0.1234,0.9876, p2: 0.8765,0.2345, r:0.5 Result: No intersection. Points further apart than circle diameter
Test: 4: p1: 0.1234,0.9876, p2: 0.1234,0.9876, r:0 Result: Radius Zero
```

## jq

Works with: jq version 1.4

In this section, a point in the plane will be represented by its Cartesian co-ordinates expressed as a JSON array: [x,y].

```# circle_centers is defined here as a filter.
# Input should be an array [x1, y1, x2, y2, r] giving the co-ordinates
# of the two points and a radius.
# If there is one solution, the output is the circle center;
# if there are two solutions centered at [x1, y1] and [x2, y2],
# then the output is [x1, y1, x2, y2];
# otherwise an explanatory string is returned.

def circle_centers:
def sq: .*.;
def c(x3; y1; y2; r; d): x3 + ((r|sq - ((d/2)|sq)) | sqrt) * (y1-y2)/d;

.[0] as \$x1 | .[1] as \$y1 | .[2] as \$x2 | .[3] as \$y2 | .[4] as \$r
| (((\$x2-\$x1)|sq) + ((\$y2-\$y1)|sq) | sqrt) as \$d
| ((\$x1+\$x2)/2) as \$x3
| ((\$y1+\$y2)/2) as \$y3
| c(\$x3; \$y1; \$y2; \$r; \$d) as \$cx1
| c(\$y3; \$x2; \$x2; \$r; \$d) as \$cy1
| (- c(-\$x3; \$y1; \$y2; \$r; \$d)) as \$cx2
| (- c(-\$y3; \$x2; \$x2; \$r; \$d)) as \$cy2
| if   \$d == 0 and \$r == 0 then [\$x1, \$y1]  # special case
elif \$d == 0     then "infinitely many circles can be drawn"
elif \$d >  \$r*2  then "points are too far from each other"
elif  0 >  \$r    then "radius is not valid"
elif (\$cx1 and \$cy1 and \$cx2 and \$cy2) | not then "no solution"
else  [\$cx1, \$cy1, \$cx2, \$cy2 ]
end;```

Examples:

```(
[0.1234,    0.9876,    0.8765,    0.2345,    2],
[0.0000,    2.0000,    0.0000,    0.0000,    1],
[0.1234,    0.9876,    0.1234,    0.9876,    2],
[0.1234,    0.9876,    0.8765,    0.2345,  0.5],
[0.1234,    0.9876,    0.1234,    0.9876,    0]
)
| "\(.) ───► \(circle_centers)"```
Output:
```\$ jq -n -c -r -f /Users/peter/jq/circle_centers.jq

[0.1234,0.9876,0.8765,0.2345,2] ───► [1.8631118016581893,1.974211801658189,-0.8632118016581896,-0.7521118016581892]
[0,2,0,0,1] ───► [0,1,0,1]
[0.1234,0.9876,0.1234,0.9876,2] ───► infinitely many circles can be drawn
[0.1234,0.9876,0.8765,0.2345,0.5] ───► points are too far from each other
[0.1234,0.9876,0.1234,0.9876,0] ───► [0.1234,0.9876]
```

## Julia

This solution uses the package AffineTransforms.jl to introduce a coordinate system (u, v) centered on the midpoint between the two points and rotated so that these points are on the u-axis. In this system, solving for the circles' centers is trivial. The two points are cast as complex numbers to aid in determining the location of the midpoint and rotation angle.

Types and Functions

```immutable Point{T<:FloatingPoint}
x::T
y::T
end

immutable Circle{T<:FloatingPoint}
c::Point{T}
r::T
end
Circle{T<:FloatingPoint}(a::Point{T}) = Circle(a, zero(T))

using AffineTransforms

function circlepoints{T<:FloatingPoint}(a::Point{T}, b::Point{T}, r::T)
cp = Circle{T}[]
r >= 0 || return (cp, "No Solution, Negative Radius")
if a == b
if abs(r) < 2eps(zero(T))
return (push!(cp, Circle(a)), "Point Solution, Zero Radius")
else
return (cp, "Infinite Solutions, Indefinite Center")
end
end
ca = Complex(a.x, a.y)
cb = Complex(b.x, b.y)
d = (ca + cb)/2
tfd = tformtranslate([real(d), imag(d)])
tfr = tformrotate(angle(cb-ca))
tfm = tfd*tfr
u = abs(cb-ca)/2
r-u > -5eps(r) || return(cp, "No Solution, Radius Too Small")
if r-u < 5eps(r)
push!(cp, Circle(apply(Point, tfm*[0.0, 0.0]), r))
return return (cp, "Single Solution, Degenerate Centers")
end
v = sqrt(r^2 - u^2)
for w in [v, -v]
push!(cp, Circle(apply(Point, tfm*[0.0, w]), r))
end
return (cp, "Two Solutions")
end
```

Main

```tp = [Point(0.1234, 0.9876),
Point(0.0000, 2.0000),
Point(0.1234, 0.9876),
Point(0.1234, 0.9876),
Point(0.1234, 0.9876)]

tq = [Point(0.8765, 0.2345),
Point(0.0000, 0.0000),
Point(0.1234, 0.9876),
Point(0.8765, 0.2345),
Point(0.1234, 0.9876)]

tr = [2.0, 1.0, 2.0, 0.5, 0.0]

println("Testing circlepoints:")
for i in 1:length(tp)
p = tp[i]
q = tq[i]
r = tr[i]
(cp, rstatus) = circlepoints(p, q, r)
println(@sprintf("(%.4f, %.4f), (%.4f, %.4f), %.4f => %s",
p.x, p.y, q.x, q.y, r, rstatus))
for c in cp
println(@sprintf("    (%.4f, %.4f), %.4f",
c.c.x, c.c.y, c.r))
end
end
```
Output:
```Testing circlepoints:
(0.1234, 0.9876), (0.8765, 0.2345), 2.0000 => Two Solutions
(1.8631, 1.9742), 2.0000
(-0.8632, -0.7521), 2.0000
(0.0000, 2.0000), (0.0000, 0.0000), 1.0000 => Single Solution, Degenerate Centers
(0.0000, 1.0000), 1.0000
(0.1234, 0.9876), (0.1234, 0.9876), 2.0000 => Infinite Solutions, Indefinite Center
(0.1234, 0.9876), (0.8765, 0.2345), 0.5000 => No Solution, Radius Too Small
(0.1234, 0.9876), (0.1234, 0.9876), 0.0000 => Point Solution, Zero Radius
(0.1234, 0.9876), 0.0000
```

## Kotlin

```// version 1.1.51

typealias IAE = IllegalArgumentException

class Point(val x: Double, val y: Double) {
fun distanceFrom(other: Point): Double {
val dx = x - other.x
val dy = y - other.y
return Math.sqrt(dx * dx + dy * dy )
}

override fun equals(other: Any?): Boolean {
if (other == null || other !is Point) return false
return (x == other.x && y == other.y)
}

override fun toString() = "(%.4f, %.4f)".format(x, y)
}

fun findCircles(p1: Point, p2: Point, r: Double): Pair<Point, Point> {
if (r < 0.0) throw IAE("the radius can't be negative")
if (r == 0.0 && p1 != p2) throw IAE("no circles can ever be drawn")
if (r == 0.0) return p1 to p1
if (p1 == p2) throw IAE("an infinite number of circles can be drawn")
val distance = p1.distanceFrom(p2)
val diameter = 2.0 * r
if (distance > diameter) throw IAE("the points are too far apart to draw a circle")
val center = Point((p1.x + p2.x) / 2.0, (p1.y + p2.y) / 2.0)
if (distance == diameter) return center to center
val mirrorDistance = Math.sqrt(r * r - distance * distance / 4.0)
val dx =  (p2.x - p1.x) * mirrorDistance / distance
val dy =  (p2.y - p1.y) * mirrorDistance / distance
return Point(center.x - dy, center.y + dx) to
Point(center.x + dy, center.y - dx)
}

fun main(args: Array<String>) {
val p = arrayOf(
Point(0.1234, 0.9876),
Point(0.8765, 0.2345),
Point(0.0000, 2.0000),
Point(0.0000, 0.0000)
)
val points = arrayOf(
p[0] to p[1], p[2] to p[3], p[0] to p[0], p[0] to p[1], p[0] to p[0]
)
val radii = doubleArrayOf(2.0, 1.0, 2.0, 0.5, 0.0)
for (i in 0..4) {
try {
val (p1, p2) = points[i]
println("For points \$p1 and \$p2 with radius \$r")
val (c1, c2) = findCircles(p1, p2, r)
if (c1 == c2)
println("there is just one circle with center at \$c1")
else
println("there are two circles with centers at \$c1 and \$c2")
}
catch(ex: IllegalArgumentException) {
println(ex.message)
}
println()
}
}
```
Output:
```For points (0.1234, 0.9876) and (0.8765, 0.2345) with radius 2.0
there are two circles with centers at (1.8631, 1.9742) and (-0.8632, -0.7521)

For points (0.0000, 2.0000) and (0.0000, 0.0000) with radius 1.0
there is just one circle with center at (0.0000, 1.0000)

For points (0.1234, 0.9876) and (0.1234, 0.9876) with radius 2.0
an infinite number of circles can be drawn

For points (0.1234, 0.9876) and (0.8765, 0.2345) with radius 0.5
the points are too far apart to draw a circle

For points (0.1234, 0.9876) and (0.1234, 0.9876) with radius 0.0
there is just one circle with center at (0.1234, 0.9876)
```

## Lambdatalk

```input:  OP1=(x1,y1), OP2=(x2,y2), r
output: OC = OH + HC
where OH = (OP1+OP2)/2
and HC = j*|HC|
where j is the unit vector rotated -90° from P1P2
and |HC| = √(r^2 - (|P1P2|/2)^2) if exists

{def circleby2points
{lambda {:x1 :y1 :x2 :y2 :r}
{if {= :r 0}
else {if {and {= :x1 :x2} {= :y1 :y2}}
then same points
else {let { {:r :r}
{:vx {- :x2 :x1}} {:vy {- :y2 :y1}}                      // v = P1P2
{:hx {/ {+ :x1 :x2} 2}} {:hy {/ {+ :y1 :y2} 2}} }        // h = OH
{let { {:r :r} {:vx :vx} {:vy :vy} {:hx :hx} {:hy :hy}          // closure
{:d {sqrt {+ {* :px :px} {* :py :py}}} } }               // d = |P1P2|
{if {> :d {* 2 :r}}                                             // d > diam
then no circle, points are too far apart
else {if {= :d {* 2 :r}}                                       // d = diam
then one circle: opposite ends of diameter with centre (:hx,:hy)
else {let { {:r :r} {:hx :hx} {:hy :hy}                        // closure
{:jx {- {/ :vy :d}}} {:jy {/ :vx :d}}              // j unit -90° to P1P2
{:d {sqrt {- {* :r :r} {/ {* :d :d} 4}}}} }        // |HC|
two circles: {br}({+ :hx {* :d :jx}},{+ :hy {* :d :jy}})  // OH + j*|HC|
{br}({- :hx {* :d :jx}},{- :hy {* :d :jy}})  // OH - j*|HC|
}}}}}}}}}

{circleby2points -1 0 1 0 0.5}
-> no circle:
points are too far apart

{circleby2points -1 0 1 0 1}
-> one circle:
opposite ends of diameter with centre (0,0)

{circleby2points -1 0 1 0 {sqrt 2}}
-> two circles:
(0,1.0000000000000002)
(0,-1.0000000000000002)

{circleby2points 0.1234 0.9876 0.8765 0.2345 2.0}
-> two circles:
(1.8631118016581893,1.974211801658189)
(-0.8632118016581896,-0.7521118016581892)

{circleby2points 0.0000 2.0000 0.0000 0.0000 1.0}
-> one circle: opposite ends of diameter with centre (0,1)

{circleby2points 0.1234 0.9876 0.1234 0.9876 2.0}
-> same points

{circleby2points 0.1234 0.9876 0.8765 0.2345 0.5}
-> no circle, points are too far apart

{circleby2points 0.1234 0.9876 0.1234 0.9876 0.0}
```

## Liberty BASIC

```'[RC] Circles of given radius through two points
for i = 1 to 5
print i;") ";x1, y1, x2, y2,r
call twoCircles x1, y1, x2, y2,r
next
end

'p1                p2           r
data 0.1234, 0.9876,    0.8765, 0.2345,    2.0
data 0.0000, 2.0000,    0.0000, 0.0000,    1.0
data 0.1234, 0.9876,    0.1234, 0.9876,    2.0
data 0.1234, 0.9876,    0.8765, 0.2345,    0.5
data 0.1234, 0.9876,    0.1234, 0.9876,    0.0

sub  twoCircles  x1, y1, x2, y2,r

if x1=x2 and y1=y2 then '2.If the points are coincident
if r=0 then ' unless r==0.0
print "It will be a single point (";x1;",";y1;") of radius 0"
exit sub
else
print "There are any number of circles via single point (";x1;",";y1;") of radius ";r
exit sub
end if
end if
r2 = sqr((x1-x2)^2+(y1-y2)^2)/2 'half distance between points
if r<r2 then
print "Points are too far apart (";2*r2;") - there are no circles of radius ";r
exit sub
end if

'else, calculate two centers
cx=(x1+x2)/2 'middle point
cy=(y1+y2)/2
'should move from middle point along perpendicular by dd2
dd2=sqr(r^2-r2^2)   'perpendicular distance
dx1=x2-cx   'vector to middle point
dy1=y2-cy
dx = 0-dy1/r2*dd2   'perpendicular:
dy = dx1/r2*dd2     'rotate and scale
print "(";cx+dy;",";cy+dx;")"   'two points, with (+)
print "(";cx-dy;",";cy-dx;")"   'and (-)

end sub```

Output:

```1) 0.1234     0.9876        0.8765        0.2345        2
(1.8631118,1.9742118)
(-0.8632118,-0.7521118)
2) 0          2             0             0             1
(0,1)
(0,1)
3) 0.1234     0.9876        0.1234        0.9876        2
There are any number of circles via single point (0.1234,0.9876) of radius 2
4) 0.1234     0.9876        0.8765        0.2345        0.5
Points are too far apart (1.06504423) - there are no circles of radius 0.5
5) 0.1234     0.9876        0.1234        0.9876        0
It will be a single point (0.1234,0.9876) of radius 0
```

## Lua

Translation of: C
```function distance(p1, p2)
local dx = (p1.x-p2.x)
local dy = (p1.y-p2.y)
return math.sqrt(dx*dx + dy*dy)
end

local seperation = distance(p1, p2)
if seperation == 0.0 then
print("No circles can be drawn through ("..p1.x..", "..p1.y..")")
else
print("Infinitely many circles can be drawn through ("..p1.x..", "..p1.y..")")
end
local cx = (p1.x+p2.x)/2
local cy = (p1.y+p2.y)/2
print("Given points are opposite ends of a diameter of the circle with center ("..cx..", "..cy..") and radius "..radius)
print("Given points are further away from each other than a diameter of a circle with radius "..radius)
else
local mirrorDistance = math.sqrt(math.pow(radius,2) - math.pow(seperation/2,2))
local dx = p2.x - p1.x
local dy = p1.y - p2.y
local ax = (p1.x + p2.x) / 2
local ay = (p1.y + p2.y) / 2
local mx = mirrorDistance * dx / seperation
local my = mirrorDistance * dy / seperation
c1 = {x=ax+my, y=ay+mx}
c2 = {x=ax-my, y=ay-mx}

print("Two circles are possible.")
end
print()
end

cases = {
{x=0.1234, y=0.9876},   {x=0.8765, y=0.2345},
{x=0.0000, y=2.0000},   {x=0.0000, y=0.0000},
{x=0.1234, y=0.9876},   {x=0.1234, y=0.9876},
{x=0.1234, y=0.9876},   {x=0.8765, y=0.2345},
{x=0.1234, y=0.9876},   {x=0.1234, y=0.9876}
}
radii = { 2.0, 1.0, 2.0, 0.5, 0.0 }
print("Case "..i)
end
```
Output:
```Case 1
Two circles are possible.
Circle C1 with center (1.8631118016582, 1.9742118016582), radius 2
Circle C2 with center (-0.86321180165819, -0.75211180165819), radius 2

Case 2
Given points are opposite ends of a diameter of the circle with center (0, 1) and radius 1

Case 3
Infinitely many circles can be drawn through (0.1234, 0.9876)

Case 4
Given points are further away from each other than a diameter of a circle with radius 0.5

Case 5
No circles can be drawn through (0.1234, 0.9876)```

## Maple

```drawCircles := proc(x1, y1, x2, y2, r, \$)
local c1, c2, p1, p2;
use geometry in
if x1 = x2 and y1 = y2 then
if r = 0 then
printf("The circle is a point at [%a, %a].\n", x1, y1);
else
printf("The two points are the same. Infinite circles can be drawn.\n");
end if;
elif evalf(distance(point(A, x1, y1), point(B, x2, y2))) >r*2 then
printf("The two points are too far apart. No circles can be drawn.\n");
else
circle(P1Cir, [A, r]);#make a circle around the first point
circle(P2Cir, [B, r]);#make a circle around the second point
intersection('i', P1Cir, P2Cir);
#the intersection of the above 2 circles should give you the centers of the two circles you need to draw
c1 := plottools[circle](coordinates(`if`(type(i, list), i[1], i)), r);#make the first circle
c2 := plottools[circle](coordinates(`if`(type(i, list), i[2], i)), r);#make the second circle
plots[display](c1, c2, scaling = constrained);#draw
end if;
end use;
end proc:

drawCircles(0.1234, 0.9876, 0.8765, 0.2345, 2.0);
drawCircles(0.0000, 2.0000, 0.0000, 0.0000, 1.0);
drawCircles(0.1234, 0.9876, 0.1234, 0.9876, 2.0);
drawCircles(0.1234, 0.9876, 0.8765, 0.2345, 0.5);
drawCircles(0.1234, 0.9876, 0.1234, 0.9876, 0.0);```
Output:
```The two points are the same. Infinite circles can be drawn.
The two points are too far apart. No circles can be drawn.
The circle is a point at [.1234, .9876].
```

## Mathematica/Wolfram Language

```Off[Solve::ratnz];
circs::equpts = "The given points (`1`, `2`) are equivalent.";
circs::dist =
"The given points (`1`, `2`) and (`3`, `4`) are too far apart for \
circs[{p1x_, p1y_}, {p1x_, p1y_}, _] :=
Message[circs::equpts, p1x, p1y];
circs[p1 : {p1x_, p1y_}, p2 : {p2x_, p2y_}, r_] /;
EuclideanDistance[p1, p2] > 2 r :=
Message[circs::dist, p1x, p1y, p2x, p2y, r];
circs[p1 : {p1x_, p1y_}, p2 : {p2x_, p2y_}, r_] :=
Values /@
Solve[Abs[x - p1x]^2 + Abs[y - p1y]^2 ==
Abs[x - p2x]^2 + Abs[y - p2y]^2 == r^2, {x, y}];
```
Output:
```circs[{.1234, .9876}, {.8765, .2345}, 2.]
{{-0.863212, -0.752112}, {1.86311, 1.97421}}
circs[{.1234, .9876}, {.1234, .9876}, 2.]
circs::equpts: The given points (0.1234`, 0.9876`) are equivalent.
circs[{.1234, .9876}, {.8765, .2345}, .5]
circs::dist: The given points (0.1234`, 0.9876`) and (0.8765`, 0.2345`) are too far apart for radius 0.5`.
circs[{.1234, .9876}, {.1234, .9876}, 0.]

## Maxima

```/* define helper function */
vabs(a):= sqrt(a.a);
realp(e):=freeof(%i, e);

/* get a general solution */
sol: block(
[p1: [x1, y1], p2: [x2, y2], c:  [x0, y0], eq],
local(r),
eq: [vabs(p1-c) = r, vabs(p2-c) = r],
assume(r>0),
args(to_poly_solve(eq, c, use_grobner = true)))\$

/* use general solution for concrete case */
getsol(sol, x1, y1, x2, y2, r):=block([n, lsol],
if [x1, y1]=[x2, y2] then (
print("infinity many solutions"),
return('infmany)),
lsol: sublist(''sol, 'realp),
n: length(lsol),
if n=0 then (
print("no solutions"),
[])
else if n=1 then (
print("single solution"),
lsol[1])
else if [assoc('x0, lsol[1]), assoc('y0, lsol[1])]=[assoc('x0, lsol[2]), assoc('y0, lsol[2])] then (
print("single solution"),
lsol[1])
else (
print("two solutions"),
lsol))\$

/* [x1, y1, x2, y2, r] */
d[1]: [0.1234, 0.9876,    0.8765, 0.2345,    2];
d[2]: [0.0000, 2.0000,    0.0000, 0.0000,    1];
d[3]: [0,      0,         0,      1,         0.4];
d[4]: [0,      0,         0,      0,         0.4];

apply('getsol, cons(sol, d[1]));
apply('getsol, cons(sol, d[2]));
apply('getsol, cons(sol, d[3]));
apply('getsol, cons(sol, d[4]));
```
Output:
```apply('getsol, cons(sol, d[1]));
two solutions
(%o9) [[x0 = 1.86311180165819, y0 = 1.974211801658189],
[x0 = - 0.86321180165819, y0 = - 0.75211180165819]]
(%i10) apply('getsol, cons(sol, d[2]));
single solution
(%o10)                       [x0 = 0.0, y0 = 1.0]
(%i11) apply('getsol, cons(sol, d[3]));
no solutions
(%o11)                                []
(%i12) apply('getsol, cons(sol, d[4]));
infinity many solutions
(%o12)                              infmany
```

## МК-61/52

```П0	С/П	П1	С/П	П2	С/П	П3	С/П	П4
ИП2	ИП0	-	x^2	ИП3	ИП1	-	x^2	+	КвКор	П5
ИП0	ИП2	+	2	/	П6	ИП1	ИП3	+	2	/	П7
ИП4	x^2	ИП5	2	/	x^2	-	КвКор	ИП5	/	П8
ИП6	ИП1	ИП3	-	ИП8	*	П9	+	ПA
ИП6	ИП9	-	ПC
ИП7	ИП2	ИП0	-	ИП8	*	П9	+	ПB
ИП7	ИП9	-	ПD
ИП5	x#0	97	8	4	ИНВ	С/П
ИП4	2	*	ИП5	-	ПE	x#0	97	ИПB	ИПA	8	5	ИНВ	С/П
ИПE	x>=0	97	8	3	ИНВ	С/П
ИПD	ИПC	ИПB	ИПA	С/П
```
Input:
``` В/О x1 С/П y1 С/П x2 С/П y2 С/П radius С/П
```
Output:
```"8.L" if the points are coincident; "8.-" if the points are opposite ends of a diameter of the circle, РY and РZ are coordinates of the center; "8.Г" if the points are farther away from each other than a diameter of a circle; else РX, РY and РZ, РT are coordinates of the circles centers.
```

## Modula-2

```MODULE Circles;
FROM EXCEPTIONS IMPORT AllocateSource,ExceptionSource,GetMessage,RAISE;
FROM FormatString IMPORT FormatString;
FROM LongMath IMPORT sqrt;
FROM LongStr IMPORT RealToStr;

VAR
TextWinExSrc : ExceptionSource;

TYPE
Point = RECORD
x,y : LONGREAL;
END;
Pair = RECORD
a,b : Point;
END;

PROCEDURE Distance(p1,p2 : Point) : LONGREAL;
VAR dx,dy : LONGREAL;
BEGIN
dx := p1.x - p2.x;
dy := p1.y - p2.y;
RETURN sqrt(dx*dx + dy*dy)
END Distance;

PROCEDURE Equal(p1,p2 : Point) : BOOLEAN;
BEGIN
RETURN (p1.x=p2.x) AND (p1.y=p2.y)
END Equal;

PROCEDURE WritePoint(p : Point);
VAR buf : ARRAY[0..63] OF CHAR;
BEGIN
WriteString("(");
RealToStr(p.x, buf);
WriteString(buf);
WriteString(", ");
RealToStr(p.y, buf);
WriteString(buf);
WriteString(")");
END WritePoint;

PROCEDURE FindCircles(p1,p2 : Point; r : LONGREAL) : Pair;
VAR
distance,diameter,mirrorDistance,dx,dy : LONGREAL;
center : Point;
BEGIN
IF r < 0.0 THEN RAISE(TextWinExSrc, 0, "the radius can't be negative") END;
IF (r = 0.0) AND NOT Equal(p1,p2) THEN RAISE(TextWinExSrc, 0, "No circles can ever be drawn") END;
IF r = 0.0 THEN RETURN Pair{p1,p1} END;
IF Equal(p1,p2) THEN RAISE(TextWinExSrc, 0, "an infinite number of circles can be drawn") END;
distance := Distance(p1,p2);
diameter := 2.0 * r;
IF distance > diameter THEN RAISE(TextWinExSrc, 0, "the points are too far apart to draw a circle") END;
center := Point{(p1.x + p2.x) / 2.0, (p1.y + p2.y) / 2.0};
IF distance = diameter THEN RETURN Pair{center, center} END;
mirrorDistance := sqrt(r * r - distance * distance / 4.0);
dx := (p2.x - p1.x) * mirrorDistance / distance;
dy := (p2.y - p1.y) * mirrorDistance / distance;
RETURN Pair{
{center.x - dy, center.y + dx},
{center.x + dy, center.y - dx}
}
END FindCircles;

PROCEDURE Print(p1,p2 : Point; r : LONGREAL) : BOOLEAN;
VAR
buf : ARRAY[0..63] OF CHAR;
result : Pair;
BEGIN
WriteString("For points ");
WritePoint(p1);
WriteString(" and ");
WritePoint(p2);
RealToStr(r, buf);
WriteString(buf);
WriteLn;

result := FindCircles(p1,p2,r);
IF Equal(result.a, result.b) THEN
WriteString("there is just one circle with the center at ");
WritePoint(result.a);
WriteLn;
ELSE
WriteString("there are two circles with centers at ");
WritePoint(result.a);
WriteString(" and ");
WritePoint(result.b);
WriteLn;
END;
WriteLn;
RETURN TRUE
EXCEPT
GetMessage(buf);
WriteString(buf);
WriteLn;
WriteLn;
RETURN FALSE
END Print;

VAR p0,p1,p2,p3 : Point;
BEGIN
AllocateSource(TextWinExSrc);
p0 := Point{0.1234,0.9876};
p1 := Point{0.8765,0.2345};
p2 := Point{0.0000,2.0000};
p3 := Point{0.0000,0.0000};

Print(p0,p1,2.0);
Print(p2,p3,1.0);
Print(p0,p0,2.0);
Print(p0,p1,0.5);
Print(p0,p0,0.0);

END Circles.
```

## Nim

Translation of: Python
```import math

type
Point = tuple[x, y: float]
Circle = tuple[x, y, r: float]

proc circles(p1, p2: Point, r: float): tuple[c1, c2: Circle] =
if r == 0: raise newException(ValueError,
if p1 == p2: raise newException(ValueError,
"coincident points gives infinite number of Circles")

# delta x, delta y between points
let (dx, dy) = (p2.x - p1.x, p2.y - p1.y)
# dist between points
let q = sqrt(dx*dx + dy*dy)
if q > 2.0*r: raise newException(ValueError,
"separation of points > diameter")

# halfway point
let p3: Point = ((p1.x+p2.x)/2, (p1.y+p2.y)/2)
# distance along the mirror line
let d = sqrt(r*r - (q/2)*(q/2))
result.c1 = (p3.x - d*dy/q, p3.y + d*dx/q, abs(r))
result.c2 = (p3.x + d*dy/q, p3.y - d*dx/q, abs(r))

const tries: seq[tuple[p1, p2: Point, r: float]] =
@[((0.1234, 0.9876), (0.8765, 0.2345), 2.0),
((0.0000, 2.0000), (0.0000, 0.0000), 1.0),
((0.1234, 0.9876), (0.1234, 0.9876), 2.0),
((0.1234, 0.9876), (0.8765, 0.2345), 0.5),
((0.1234, 0.9876), (0.1234, 0.9876), 0.0)]

for p1, p2, r in tries.items:
echo "Through points:"
echo "  ", p1
echo "  ", p2
echo "  and radius ", r
echo "You can construct the following circles:"
try:
let (c1, c2) = circles(p1, p2, r)
echo "  ", c1
echo "  ", c2
except ValueError:
echo "  ERROR: ", getCurrentExceptionMsg()
echo ""
```
Output:
```Through points:
(x: 0.1234, y: 0.9876)
(x: 0.8764999999999999, y: 0.2345)
You can construct the following circles:
(x: 1.863111801658189, y: 1.974211801658189, r: 2.0)
(x: -0.8632118016581896, y: -0.7521118016581892, r: 2.0)

Through points:
(x: 0.0, y: 2.0)
(x: 0.0, y: 0.0)
You can construct the following circles:
(x: 0.0, y: 1.0, r: 1.0)
(x: 0.0, y: 1.0, r: 1.0)

Through points:
(x: 0.1234, y: 0.9876)
(x: 0.1234, y: 0.9876)
You can construct the following circles:
ERROR: coincident points gives infinite number of Circles

Through points:
(x: 0.1234, y: 0.9876)
(x: 0.8764999999999999, y: 0.2345)
You can construct the following circles:
ERROR: separation of points > diameter

Through points:
(x: 0.1234, y: 0.9876)
(x: 0.1234, y: 0.9876)
You can construct the following circles:

## OCaml

Original version by User:Vanyamil

```(* Task : Circles of given radius through two points *)

(* Types to make code even more readable *)
type point = float * float
type circle = Circle of radius * point
type circ_output =
NoSolution
| OneSolution of circle
| TwoSolutions of circle * circle
| InfiniteSolutions
;;

(* Actual function *)
let circles_2points_radius (x1, y1 : point) (x2, y2 : point) (r : radius) =
let (dx, dy) = (x2 -. x1, y2 -. y1) in
let dist_sq = dx *. dx +. dy *. dy in
match dist_sq, r with
(* Edge case - point circles *)
| 0., 0. -> OneSolution (Circle (r, (x1, y1)))
(* Edge case - coinciding points *)
| 0., _ -> InfiniteSolutions
| _ ->
let side_len_sq = r *. r -. dist_sq /. 4. in
let midp = ((x1 +. x2) *. 0.5, (y1 +. y2) *. 0.5) in
(* Points are too far apart; same whether r = 0 or not *)
if side_len_sq < 0. then NoSolution
(* Points are on diameter *)
else if side_len_sq = 0. then OneSolution (Circle (r, midp))
else
(* A right-angle triangle is made with the radius as hyp, dist/2 as one side *)
let side_len = sqrt (r *. r -. dist_sq /. 4.) in
let dist = sqrt dist_sq in
(*  A 90-deg rotation of a vector (x, y) is obtained by either (y, -x) or (-y, x)
We need both, so pick one and the other is its negative.
*)
let (vx, vy) = (-. dy *. side_len /. dist, dx *. side_len /. dist) in
let c1 = Circle (r, (fst midp +. vx, snd midp +. vy)) in
let c2 = Circle (r, (fst midp -. vx, snd midp -. vy)) in
TwoSolutions (c1, c2)
;;

(* Relevant tests and printing *)
let tests = [
(0.1234, 0.9876), (0.8765, 0.2345), 2.0;
(0.0000, 2.0000), (0.0000, 0.0000), 1.0;
(0.1234, 0.9876), (0.1234, 0.9876), 2.0;
(0.1234, 0.9876), (0.8765, 0.2345), 0.5;
(0.1234, 0.9876), (0.1234, 0.9876), 0.0;
] ;;

let format_output (out : circ_output) = match out with
| NoSolution -> print_endline "No solution"
| OneSolution (Circle (_, (x, y))) -> Printf.printf "One solution: (%.6f, %.6f)\n" x y
| TwoSolutions (Circle (_, (x1, y1)), Circle (_, (x2, y2))) ->
Printf.printf "Two solutions: (%.6f, %.6f) and (%.6f, %.6f)\n" x1 y1 x2 y2
| InfiniteSolutions -> print_endline "Infinite solutions"
;;

let _ =
List.iter
(fun (a, b, c) -> circles_2points_radius a b c |> format_output)
tests
;;
```
Output:
```Two solutions: (1.863112, 1.974212) and (-0.863212, -0.752112)
One solution: (0.000000, 1.000000)
Infinite solutions
No solution
One solution: (0.123400, 0.987600)
```

## Oforth

```: circleCenter(x1, y1, x2, y2, r)
| d xmid ymid r1 md |
x2 x1 - sq  y2 y1 - sq + sqrt -> d
x1 x2 + 2 / -> xmid
y1 y2 + 2 / -> ymid
2 r * -> r1

d 0.0 == ifTrue: [ "Infinite number of circles" . return ]
d r1 == ifTrue:  [ System.Out "One circle: (" << xmid << ", " << ymid << ")" << cr return ]
d r1  > ifTrue:  [ "No circle" . return ]

r sq d 2 / sq - sqrt ->md

System.Out "C1 : (" << xmid y1 y2 - md * d / + << ", " << ymid x2 x1 - md * d / + << ")" << cr
System.Out "C2 : (" << xmid y1 y2 - md * d / - << ", " << ymid x2 x1 - md * d / - << ")" << cr
;```
Output:
```
>0.1234 0.9876 0.8765 0.2345 2 circleCenter
C1 : (1.86311180165819, 1.97421180165819)
C2 : (-0.86321180165819, -0.752111801658189)
ok

>0 2 0 0 1 circleCenter
One cirlce: (0, 1)
ok

>0.1234 0.9876 0.8765 0.2345 0.5 circleCenter
No circle ok

>0.1234 0.9876 0.1234 0.9876 0 circleCenter
Infinite number of circles ok

```

## ooRexx

Translation of: REXX
```/*REXX pgm finds 2 circles with a specific radius given two (X,Y) points*/
a.=''
a.1=0.1234 0.9876 0.8765 0.2345 2
a.2=0.0000 2.0000 0.0000 0.0000 1
a.3=0.1234 0.9876 0.1234 0.9876 2
a.4=0.1234 0.9876 0.8765 0.2345 0.5
a.5=0.1234 0.9876 0.1234 0.9876 0

Say '     x1      y1      x2      y2  radius   cir1x   cir1y   cir2x   cir2y'
Say ' ------  ------  ------  ------  ------  ------  ------  ------  ------'
Do j=1 By 1 While a.j<>''
Do k=1 For 4
w.k=f(word(a.j,k))
End
Say w.1 w.2 w.3 w.4 format(word(a.j,5),5,1)  twocircles(a.j)
End
Exit

twocircles: Procedure
Parse Arg px py qx qy r .
If r=0 Then
Return ' radius of zero gives no circles.'
x=(qx-px)/2
y=(qy-py)/2
bx=px+x
by=py+y
pb=rxCalcsqrt(x**2+y**2)
If pb=0 Then
Return ' coincident points give infinite circles'
If pb>r Then
Return ' points are too far apart for the given radius'
cb=rxCalcsqrt(r**2-pb**2)
x1=y*cb/pb
y1=x*cb/pb
Return f(bx-x1) f(by+y1) f(bx+x1) f(by-y1)

f: Return format(arg(1),2,4) /* format a number with 4 dec dig.*/

::requires 'rxMath' library```
Output:
```     x1      y1      x2      y2  radius   cir1x   cir1y   cir2x   cir2y
------  ------  ------  ------  ------  ------  ------  ------  ------
0.1234  0.9876  0.8765  0.2345     2.0  1.8631  1.9742 -0.8632 -0.7521
0.0000  2.0000  0.0000  0.0000     1.0  0.0000  1.0000  0.0000  1.0000
0.1234  0.9876  0.1234  0.9876     2.0  coincident points give infinite circles
0.1234  0.9876  0.8765  0.2345     0.5  points are too far apart for the given radius
0.1234  0.9876  0.1234  0.9876     0.0  radius of zero gives no circles.```

```// distance between two points
function distance(p1, p2) = sqrt((difference(p2.x, p1.x)) ^ 2 + (difference(p2.y, p1.y) ^ 2));
// difference between two values in any order
function difference(a, b) = let(x = a > b ? a - b : b - a) x;

// function to find the circles of given radius through two points
let(mid = (p1 + p2)/2, q = distance(p1, p2), x_dist = sqrt(radius ^ 2 - (q / 2) ^ 2) * (p1.y - p2.y) / q,
y_dist = sqrt(radius ^ 2 - (q / 2) ^ 2) * (p2.x - p1.x) / q)
// point 1 and point 2 must not be the same point
assert(p1 != p2)
// radius must be more than 0
// distance between points cannot be more than diameter
// return both qualifying centres
[mid + [ x_dist, y_dist ], mid - [ x_dist, y_dist ]];

{
tests = [
[ [ -10, -10, 0 ], [ 50, 0, 0 ], 100 ], [ [ 200, 0, 0 ], [ 220, -20, 0 ], 30 ],
[ [ 300, 100, 0 ], [ 350, 200, 0 ], 80 ]
];

for (t = tests)
{
let(start = t[0], end = t[1], radius = t[2])
{
// plot start and end dots - these should be at the intersections of the circles
color("green") translate(start) cylinder(h = 3, r = 4);
color("green") translate(end) cylinder(h = 3, r = 4);

// call function
echo("centres", centres);
// plot results
color("yellow") translate(centres[0]) cylinder(h = 1, r = radius);
color("red") translate(centres[1]) cylinder(h = 2, r = radius);
};
};
// The following tests will stop all execution. To run them, uncomment one at a time
// should fail - same points
// should fail - points are more than diameter apart
// should fail - radius must be greater than 0
}

```

## PARI/GP

```circ(a, b, r)={
if(a==b, return("impossible"));
my(h=(b-a)/2,t=sqrt(r^2-abs(h)^2)/abs(h)*h);
[a+h+t*I,a+h-t*I]
};
circ(0.1234 + 0.9876*I, 0.8765 + 0.2345*I, 2)
circ(0.0000 + 2.0000*I, 0.0000 + 0.0000*I, 1)
circ(0.1234 + 0.9876*I, 0.1234 + 0.9876*I, 2)
circ(0.1234 + 0.9876*I, 0.8765 + 0.2345*I, .5)
circ(0.1234 + 0.9876*I, 0.1234 + 0.9876*I, 0)```
Output:
```%1 = [1.86311180 + 1.97421180*I, -0.863211802 - 0.752111802*I]
%2 = [0.E-9 + 1.00000000*I, 0.E-9 + 1.00000000*I]
%3 = "impossible"
%4 = [0.370374144 + 0.740625856*I, 0.629525856 + 0.481474144*I]
%5 = "impossible"```

## Perl

Translation of: Python
```use strict;

sub circles {
my (\$x1, \$y1, \$x2, \$y2, \$r) = @_;

return "Radius is zero" if \$r == 0;
return "Coincident points gives infinite number of circles" if \$x1 == \$x2 and \$y1 == \$y2;

# delta x, delta y between points
my (\$dx, \$dy) = (\$x2 - \$x1, \$y2 - \$y1);
my \$q = sqrt(\$dx**2 + \$dy**2);
return "Separation of points greater than diameter" if \$q > 2*\$r;

# halfway point
my (\$x3, \$y3) = ((\$x1 + \$x2) / 2, (\$y1 + \$y2) / 2);
# distance along the mirror line
my \$d = sqrt(\$r**2-(\$q/2)**2);

# pair of solutions
sprintf '(%.4f, %.4f) and (%.4f, %.4f)',
\$x3 - \$d*\$dy/\$q, \$y3 + \$d*\$dx/\$q,
\$x3 + \$d*\$dy/\$q, \$y3 - \$d*\$dx/\$q;
}

my @arr = (
[0.1234, 0.9876, 0.8765, 0.2345, 2.0],
[0.0000, 2.0000, 0.0000, 0.0000, 1.0],
[0.1234, 0.9876, 0.1234, 0.9876, 2.0],
[0.1234, 0.9876, 0.8765, 0.2345, 0.5],
[0.1234, 0.9876, 0.1234, 0.9876, 0.0]
);

printf "(%.4f, %.4f) and (%.4f, %.4f) with radius %.1f: %s\n", @\$_[0..4], circles @\$_ for @arr;
```
Output:
```(0.1234, 0.9876) and (0.8765, 0.2345) with radius 2.0: (1.8631, 1.9742) and (-0.8632, -0.7521)
(0.0000, 2.0000) and (0.0000, 0.0000) with radius 1.0: (0.0000, 1.0000) and (0.0000, 1.0000)
(0.1234, 0.9876) and (0.1234, 0.9876) with radius 2.0: Coincident points gives infinite number of circles
(0.1234, 0.9876) and (0.8765, 0.2345) with radius 0.5: Separation of points greater than diameter

## Phix

```with javascript_semantics
constant tests = {{0.1234, 0.9876, 0.8765, 0.2345, 2.0},
{0.0000, 2.0000, 0.0000, 0.0000, 1.0},
{0.1234, 0.9876, 0.1234, 0.9876, 2.0},
{0.1234, 0.9876, 0.8765, 0.2345, 0.5},
{0.1234, 0.9876, 0.1234, 0.9876, 0.0}}
for i=1 to length(tests) do
atom {x1,y1,x2,y2,r} = tests[i],
xd = x2-x1, yd = y1-y2,
s2 = xd*xd+yd*yd, sep = sqrt(s2),
xh = (x1+x2)/2, yh = (y1+y2)/2
string txt
if sep=0 then
txt = "same points/"&iff(r=0?"radius is zero":"infinite solutions")
elsif sep=2*r then
txt = sprintf("opposite ends of diameter with centre {%.4f,%.4f}",{xh,yh})
elsif sep>2*r then
txt = sprintf("too far apart (%.4f > %.4f)",{sep,2*r})
else
atom md = sqrt(r*r-s2/4), xs = md*xd/sep, ys = md*yd/sep
txt = sprintf("{%.4f,%.4f} and {%.4f,%.4f}",{xh+ys,yh+xs,xh-ys,yh-xs})
end if
printf(1,"points {%.4f,%.4f}, {%.4f,%.4f} with radius %.1f ==> %s\n",{x1,y1,x2,y2,r,txt})
end for
```
Output:
```points {0.1234,0.9876}, {0.8765,0.2345} with radius 2.0 ==> {1.8631,1.9742} and {-0.8632,-0.7521}
points {0.0000,2.0000}, {0.0000,0.0000} with radius 1.0 ==> opposite ends of diameter with centre {0.0000,1.0000}
points {0.1234,0.9876}, {0.1234,0.9876} with radius 2.0 ==> same points/infinite solutions
points {0.1234,0.9876}, {0.8765,0.2345} with radius 0.5 ==> too far apart (1.0650 > 1.0000)
```

## PL/I

Translation of: REXX
```twoci: Proc Options(main);
Dcl 1 *(5),
2 m1x Dec Float Init(0.1234,     0,0.1234,0.1234,0.1234),
2 m1y Dec Float Init(0.9876,     2,0.9876,0.9876,0.9876),
2 m2x Dec Float Init(0.8765,     0,0.1234,0.8765,0.1234),
2 m2y Dec Float Init(0.2345,     0,0.9876,0.2345,0.9876),
2 r   Dec Float Init(     2,     1,     2,0.5   ,     0);
Dcl i Bin Fixed(31);
Put Edit('     x1     y1     x2     y2  r '||
'  cir1x   cir1y   cir2x   cir2y')(Skip,a);
Put Edit(' ====== ====== ====== ======  = '||
' ======  ======  ======  ======')(Skip,a);
Do i=1 To 5;
Put Edit(m1x(i),m1y(i),m2x(i),m2y(i),r(i))
(Skip,4(f(7,4)),f(3));
Put Edit(twocircles(m1x(i),m1y(i),m2x(i),m2y(i),r(i)))(a);
End;

twoCircles: proc(m1x,m1y,m2x,m2y,r) Returns(Char(50) Var);
Dcl (m1x,m1y,m2x,m2y,r) Dec Float;
Dcl (cx,cy,bx,by,pb,x,y,x1,y1) Dec Float;
Dcl res Char(50) Var;
If r=0 then return(' radius of zero gives no circles.');
x=(m2x-m1x)/2;
y=(m2y-m1y)/2;
bx=m1x+x;
by=m1y+y;
pb=sqrt(x**2+y**2);
cx=(m2x-m1x)/2;
cy=(m2y-m1y)/2;
bx=m1x+x;
by=m1y+y;
pb=sqrt(x**2+y**2)
if pb=0 then return(' coincident points give infinite circles');
if pb>r then return(' points are too far apart for the given radius');
cb=sqrt(r**2-pb**2);
x1=y*cb/pb;
y1=x*cb/pb
Put String(res) Edit((bx-x1),(by+y1),(bx+x1),(by-y1))(4(f(8,4)));
Return(res);
End;
End;```
Output:
```     x1     y1     x2     y2  r   cir1x   cir1y   cir2x   cir2y
====== ====== ====== ======  =  ======  ======  ======  ======
0.1234 0.9876 0.8765 0.2345  2  1.8631  1.9742 -0.8632 -0.7521
0.0000 2.0000 0.0000 0.0000  1  0.0000  1.0000  0.0000  1.0000
0.1234 0.9876 0.1234 0.9876  2 coincident points give infinite circles
0.1234 0.9876 0.8765 0.2345  1 points are too far apart for the given radius
0.1234 0.9876 0.1234 0.9876  0 radius of zero gives no circles.
```

## PureBasic

```DataSection
DataStart:
Data.d  0.1234, 0.9876,   0.8765, 0.2345,   2.0
Data.d  0.0000, 2.0000,   0.0000, 0.0000,   1.0
Data.d  0.1234, 0.9876,   0.1234, 0.9876,   2.0
Data.d  0.1234, 0.9876,   0.9765, 0.2345,   0.5
Data.d  0.1234, 0.9876,   0.1234, 0.9876,   0.0
DataEnd:
EndDataSection
Macro MaxRec : (?DataEnd-?DataStart)/SizeOf(P2r)-1 : EndMacro

Structure Pxy   : x.d         : y.d           : EndStructure
Structure P2r   : p1.Pxy      : p2.Pxy  : r.d : EndStructure
Structure PData : Prec.P2r[5]                 : EndStructure

Procedure.s cCenter(Rec.i)
If Rec<0 Or Rec>MaxRec : ProcedureReturn "Data set number incorrect." : EndIf
*myP.PData=?DataStart
r.d=*myP\Prec[Rec]\r
If r<=0.0 : ProcedureReturn "Illegal radius." : EndIf
r2.d=2.0*r
x1.d=*myP\Prec[Rec]\p1\x : x2.d=*myP\Prec[Rec]\p2\x
y1.d=*myP\Prec[Rec]\p1\y : y2.d=*myP\Prec[Rec]\p2\y
d.d=Sqr(Pow(x2-x1,2)+Pow(y2-y1,2))
If d=0.0 : ProcedureReturn "Identical points, infinite number of circles." : EndIf
If d>r2  : ProcedureReturn "No circles possible." : EndIf
z.d=Sqr(Pow(r,2)-Pow(d/2.0,2))
x3.d =(x1+x2)/2.0     :   y3.d =(y1+y2)/2.0
cx1.d=x3+z*(y1-y2)/d  :   cy1.d=y3+z*(x2-x1)/d
cx2.d=x3-z*(y1-y2)/d  :   cy2.d=y3-z*(x2-x1)/d
If d=r2 : ProcedureReturn "Single circle at ("+StrD(cx1)+","+StrD(cy1)+")" : EndIf
ProcedureReturn "("+StrD(cx1)+","+StrD(cy1)+") and ("+StrD(cx2)+","+StrD(cy2)+")"
EndProcedure

If OpenConsole("")
For i=0 To MaxRec : PrintN(cCenter(i)) : Next : Input()
EndIf```
Output:
```(1.8631118017,1.9742118017) and (-0.8632118017,-0.7521118017)
Single circle at (0,1)
Identical points, infinite number of circles.
No circles possible.

## Python

The function raises the ValueError exception for the special cases and uses try - except to catch these and extract the exception detail.

```from collections import namedtuple
from math import sqrt

Pt = namedtuple('Pt', 'x, y')
Circle = Cir = namedtuple('Circle', 'x, y, r')

def circles_from_p1p2r(p1, p2, r):
'Following explanation at http://mathforum.org/library/drmath/view/53027.html'
if r == 0.0:
(x1, y1), (x2, y2) = p1, p2
if p1 == p2:
raise ValueError('coincident points gives infinite number of Circles')
# delta x, delta y between points
dx, dy = x2 - x1, y2 - y1
# dist between points
q = sqrt(dx**2 + dy**2)
if q > 2.0*r:
raise ValueError('separation of points > diameter')
# halfway point
x3, y3 = (x1+x2)/2, (y1+y2)/2
# distance along the mirror line
d = sqrt(r**2-(q/2)**2)
c1 = Cir(x = x3 - d*dy/q,
y = y3 + d*dx/q,
r = abs(r))
c2 = Cir(x = x3 + d*dy/q,
y = y3 - d*dx/q,
r = abs(r))
return c1, c2

if __name__ == '__main__':
for p1, p2, r in [(Pt(0.1234, 0.9876), Pt(0.8765, 0.2345), 2.0),
(Pt(0.0000, 2.0000), Pt(0.0000, 0.0000), 1.0),
(Pt(0.1234, 0.9876), Pt(0.1234, 0.9876), 2.0),
(Pt(0.1234, 0.9876), Pt(0.8765, 0.2345), 0.5),
(Pt(0.1234, 0.9876), Pt(0.1234, 0.9876), 0.0)]:
print('Through points:\n  %r,\n  %r\n  and radius %f\nYou can construct the following circles:'
% (p1, p2, r))
try:
print('  %r\n  %r\n' % circles_from_p1p2r(p1, p2, r))
except ValueError as v:
print('  ERROR: %s\n' % (v.args[0],))
```
Output:
```Through points:
Pt(x=0.1234, y=0.9876),
Pt(x=0.8765, y=0.2345)
You can construct the following circles:
Circle(x=1.8631118016581893, y=1.974211801658189, r=2.0)
Circle(x=-0.8632118016581896, y=-0.7521118016581892, r=2.0)

Through points:
Pt(x=0.0, y=2.0),
Pt(x=0.0, y=0.0)
You can construct the following circles:
Circle(x=0.0, y=1.0, r=1.0)
Circle(x=0.0, y=1.0, r=1.0)

Through points:
Pt(x=0.1234, y=0.9876),
Pt(x=0.1234, y=0.9876)
You can construct the following circles:
ERROR: coincident points gives infinite number of Circles

Through points:
Pt(x=0.1234, y=0.9876),
Pt(x=0.8765, y=0.2345)
You can construct the following circles:
ERROR: separation of points > diameter

Through points:
Pt(x=0.1234, y=0.9876),
Pt(x=0.1234, y=0.9876)
You can construct the following circles:

## Racket

Using library `plot/utils` for simple vector operations.

```#lang racket
(require plot/utils)

(define (circle-centers p1 p2 r)
(when (zero? r) (err "zero radius."))
(when (equal? p1 p2) (err "the points coinside."))
; the midle point
(define m (v/ (v+ p1 p2) 2))
; the vector connecting given points
(define d (v/ (v- p1 p2) 2))
; the distance between the center of the circle and the middle point
(define ξ (- (sqr r) (vmag^2 d)))
(when (negative? ξ) (err "given radius is less then the distance between points."))
; the unit vector orthogonal to the delta
(define n (vnormalize (orth d)))
; the shift along the direction orthogonal to the delta
(define x (v* n (sqrt ξ)))
(values (v+ m x) (v- m x)))

;; error message
(define (err m) (error "Impossible to build a circle:" m))

;; returns a vector which is orthogonal to the geven one
(define orth (match-lambda [(vector x y) (vector y (- x))]))
```
Testing:
```> (circle-centers #(0.1234 0.9876) #(0.8765 0.2345) 2.0)
'#(1.8631118016581893 1.974211801658189)
'#(-0.8632118016581896 -0.7521118016581892)

> (circle-centers #(0.0000 2.0000) #(0.0000 0.0000) 1.0)
'#(0.0 1.0)
'#(0.0 1.0)

> (circle-centers #(0.1234 0.9876) #(0.1234 0.9876) 2.0)
. . Impossible to find a circle: "the points coinside."

> (circle-centers #(0.1234 0.9876) #(0.8765 0.2345) 0.5)
. . Impossible to find a circle: "given radius is less then the distance between points."

> (circle-centers #(0.1234 0.9876) #(0.1234 0.9876) 0.0)
. . Impossible to find a circle: "zero radius."
```

Drawing circles:

```(require 2htdp/image)

(define/match (point v)
[{(vector x y)} (λ (s) (place-image (circle 2 "solid" "black") x y s))])

(define/match (circ v r)
[{(vector x y) r} (λ (s) (place-image (circle r "outline" "red") x y s))])

(define p1 #(40 50))
(define p2 #(60 30))
(define r 20)
(define-values (x1 x2) (circle-centers p1 p2 r))

((compose (point p1) (point p2) (circ x1 r) (circ x2 r))
(empty-scene 100 100))
```

## Raku

(formerly Perl 6)

Works with: Rakudo version 2020.08.1
```multi sub circles (@A, @B where ([and] @A Z== @B), 0.0) { 'Degenerate point' }
multi sub circles (@A, @B where ([and] @A Z== @B), \$)   { 'Infinitely many share a point' }
multi sub circles (@A, @B, \$radius) {
my @middle = (@A Z+ @B) X/ 2;
my @diff = @A Z- @B;
my \$q = sqrt [+] @diff X** 2;
return 'Too far apart' if \$q > \$radius * 2;

my @orth = -@diff[0], @diff[1] X* sqrt(\$radius ** 2 - (\$q / 2) ** 2) / \$q;
return (@middle Z+ @orth), (@middle Z- @orth);
}

my @input =
([0.1234, 0.9876],  [0.8765, 0.2345],   2.0),
([0.0000, 2.0000],  [0.0000, 0.0000],   1.0),
([0.1234, 0.9876],  [0.1234, 0.9876],   2.0),
([0.1234, 0.9876],  [0.8765, 0.2345],   0.5),
([0.1234, 0.9876],  [0.1234, 0.9876],   0.0),
;

for @input {
say .list.raku, ': ', circles(|\$_).join(' and ');
}
```
Output:
```([0.1234, 0.9876], [0.8765, 0.2345], 2.0): 1.86311180165819 1.97421180165819 and -0.863211801658189 -0.752111801658189
([0.0, 2.0], [0.0, 0.0], 1.0): 0 1 and 0 1
([0.1234, 0.9876], [0.1234, 0.9876], 2.0): Infinitely many share a point
([0.1234, 0.9876], [0.8765, 0.2345], 0.5): Too far apart
([0.1234, 0.9876], [0.1234, 0.9876], 0.0): Degenerate point```

Another possibility is to use the Complex plane, for it often makes calculations easier with plane geometry:

```multi sub circles (\$a, \$b where \$a == \$b, 0.0) { 'Degenerate point' }
multi sub circles (\$a, \$b where \$a == \$b, \$)   { 'Infinitely many share a point' }
multi sub circles (\$a, \$b, \$r) {
my \$h = (\$b - \$a) / 2;
my \$l = sqrt(\$r**2 - \$h.abs**2);
return 'Too far apart' if \$l.isNaN;
return map { \$a + \$h + \$l * \$_ * \$h / \$h.abs }, i, -i;
}

my @input =
(0.1234 + 0.9876i,  0.8765 + 0.2345i,   2.0),
(0.0000 + 2.0000i,  0.0000 + 0.0000i,   1.0),
(0.1234 + 0.9876i,  0.1234 + 0.9876i,   2.0),
(0.1234 + 0.9876i,  0.8765 + 0.2345i,   0.5),
(0.1234 + 0.9876i,  0.1234 + 0.9876i,   0.0),
;

for @input {
say .join(', '), ': ', circles(|\$_).join(' and ');
}
```
Output:
```0.1234+0.9876i, 0.8765+0.2345i, 2: 1.86311180165819+1.97421180165819i and -0.863211801658189-0.752111801658189i
0+2i, 0+0i, 1: 0+1i and 0+1i
0.1234+0.9876i, 0.1234+0.9876i, 2: Infinitely many share a point
0.1234+0.9876i, 0.8765+0.2345i, 0.5: Too far apart
0.1234+0.9876i, 0.1234+0.9876i, 0: Degenerate point```

## REXX

Translation of: XPL0

The REXX language doesn't have a   sqrt   function,   so one is included below.

```/*REXX pgm finds 2 circles with a specific radius given 2 (X1,Y1) and (X2,Y2) ctr points*/
@.=; @.1= 0.1234   0.9876    0.8765    0.2345    2
@.2= 0        2         0         0         1
@.3= 0.1234   0.9876    0.1234    0.9876    2
@.4= 0.1234   0.9876    0.8765    0.2345    0.5
@.5= 0.1234   0.9876    0.1234    0.9876    0
say '     x1        y1        x2        y2     radius          circle1x  circle1y  circle2x  circle2y'
say '  ════════  ════════  ════════  ════════  ══════          ════════  ════════  ════════  ════════'
do  j=1  while  @.j\=='';  parse var @.j  p1 p2 p3 p4 r           /*points, radii*/
say fmt(p1)  fmt(p2)  fmt(p3)  fmt(p4)    center(r/1, 9)    "───► "      2circ(@.j)
end   /*j*/
exit 0                                           /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
2circ: procedure; parse arg px py qx qy r .;     x= (qx-px)/2;          y= (qy-py)/2
bx= px + x;            by= py + y
pb= sqrt(x**2 + y**2)
if r = 0  then return  'radius of zero yields no circles.'
if pb==0  then return  'coincident points give infinite circles.'
if pb >r  then return  'points are too far apart for the specified radius.'
cb= sqrt(r**2 - pb**2);     x1= y * cb / pb;                       y1= x * cb / pb
return  fmt(bx-x1)   fmt(by+y1)   fmt(bx+x1)   fmt(by-y1)
/*──────────────────────────────────────────────────────────────────────────────────────*/
fmt:   arg f;   f= right( format(f, , 4), 9);         _= f  /*format # with 4 dec digits*/
if pos(.,f)>0 & pos('E',f)=0  then f= strip(f,'T',0) /*strip trailing 0s if .& ¬E*/
return left( strip(f, 'T', .), length(_) )           /*strip trailing dec point. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
sqrt:  procedure; arg x; if x=0  then return 0;  d=digits(); numeric digits;  h=d+6;  m.=9
numeric form;  parse value format(x,2,1,,0) 'E0'  with  g "E" _ .;  g=g *.5'e'_ % 2
do j=0  while h>9;      m.j=h;               h=h%2+1;       end  /*j*/
do k=j+5  to 0  by -1;  numeric digits m.k;  g=(g+x/g)*.5;  end  /*k*/;  return g
```
output   when using the default inputs:
```     x1        y1        x2        y2     radius          circle1x  circle1y  circle2x  circle2y
════════  ════════  ════════  ════════  ══════          ════════  ════════  ════════  ════════
0.1234    0.9876    0.8765    0.2345     2     ───►     1.8631    1.9742   -0.8632   -0.7521
0         2         0         0          1     ───►     0         1         0         1
0.1234    0.9876    0.1234    0.9876     2     ───►  coincident points give infinite circles.
0.1234    0.9876    0.8765    0.2345    0.5    ───►  points are too far apart for the given radius.
0.1234    0.9876    0.1234    0.9876     0     ───►  radius of zero gives no circles.
```

## Ring

```# Project : Circles of given radius through two points

decimals(4)
x1 = 0.1234
y1 = 0.9876
x2 = 0.8765
y2 = 0.2345
r = 2.0
see "1 : " + x1 + " " + y1 + " " + x2 + " " + y2 + " " + r + nl
twocircles(x1, y1, x2, y2, r)

x1 = 0.0000
y1 = 2.0000
x2 = 0.0000
y2 = 0.0000
r = 1.0
see "2 : " + x1 + " " + y1 + " " + x2 + " " + y2 + " " + r + nl
twocircles(x1, y1, x2, y2, r)

x1 = 0.1234
y1 = 0.9876
x2 = 0.1234
y2 = 0.9876
r = 2.0
see "3 : " + x1 + " " + y1 + " " + x2 + " " + y2 + " " + r + nl
twocircles(x1, y1, x2, y2, r)

x1 = 0.1234
y1 = 0.9876
x2 = 0.8765
y2 = 0.2345
r = 0.5
see "4 : " + x1 + " " + y1 + " " + x2 + " " + y2 + " " + r + nl
twocircles(x1, y1, x2, y2, r)

x1 = 0.1234
y1 = 0.9876
x2 = 0.1234
y2 = 0.9876
r= 0.0
see "5 : " + x1 + " " + y1 + " " + x2 + " " + y2 + " " + r + nl
twocircles(x1, y1, x2, y2, r)

func twocircles(x1, y1, x2, y2, r)
if x1=x2 and y1=y2
if r=0
see "It will be a single point (" + x1 + "," + y1 + ") of radius 0" + nl + nl
return
else
see "There are any number of circles via single point (" + x1 + "," + y1 + ") of radius " + r + nl + nl
return
ok
ok
r2 = sqrt(pow((x1-x2),2)+pow((y1-y2),2))/2
if r<r2
see "Points are too far apart (" + 2*r2 + ") - there are no circles of radius " + r + nl + nl
return
ok
cx=(x1+x2)/2
cy=(y1+y2)/2
dd2=sqrt(pow(r,2)-pow(r2,2))
dx1=x2-cx
dy1=y2-cy
dx = 0-dy1/r2*dd2
dy = dx1/r2*dd2
see "(" + (cx+dy) + ", " + (cy+dx) + ")" + nl
see "(" + (cx-dy) + ", " + (cy-dx) + ")" + nl + nl```

Output:

```1 : 0.1234 0.9876 0.8765 0.2345 2
(1.8631, 1.9742)
(-0.8632, -0.7521)

2 : 0 2 0 0 1
(0, 1)
(0, 1)

3 : 0.1234 0.9876 0.1234 0.9876 2
There are any number of circles via single point (0.1234,0.9876) of radius 2

4 : 0.1234 0.9876 0.8765 0.2345 0.5000
Points are too far apart (1.0650) - there are no circles of radius 0.5000

5 : 0.1234 0.9876 0.1234 0.9876 0
It will be a single point (0.1234,0.9876) of radius 0
```

## Ruby

Translation of: Python
```Pt     = Struct.new(:x, :y)
Circle = Struct.new(:x, :y, :r)

def circles_from(pt1, pt2, r)
raise ArgumentError, "Infinite number of circles, points coincide." if pt1 == pt2 && r > 0
# handle single point and r == 0
return [Circle.new(pt1.x, pt1.y, r)] if pt1 == pt2 && r == 0
dx, dy = pt2.x - pt1.x, pt2.y - pt1.y
# distance between points
q = Math.hypot(dx, dy)
# Also catches pt1 != pt2 && r == 0
raise ArgumentError, "Distance of points > diameter." if q > 2.0*r
# halfway point
x3, y3 = (pt1.x + pt2.x)/2.0, (pt1.y + pt2.y)/2.0
d = (r**2 - (q/2)**2)**0.5
[Circle.new(x3 - d*dy/q, y3 + d*dx/q, r),
Circle.new(x3 + d*dy/q, y3 - d*dx/q, r)].uniq
end

# Demo:
ar = [[Pt.new(0.1234, 0.9876), Pt.new(0.8765, 0.2345), 2.0],
[Pt.new(0.0000, 2.0000), Pt.new(0.0000, 0.0000), 1.0],
[Pt.new(0.1234, 0.9876), Pt.new(0.1234, 0.9876), 2.0],
[Pt.new(0.1234, 0.9876), Pt.new(0.8765, 0.2345), 0.5],
[Pt.new(0.1234, 0.9876), Pt.new(0.1234, 0.9876), 0.0]]

ar.each do |p1, p2, r|
print "Given points:\n  #{p1.values},\n  #{p2.values}\n  and radius #{r}\n"
begin
circles = circles_from(p1, p2, r)
puts "You can construct the following circles:"
circles.each{|c| puts "  #{c}"}
rescue ArgumentError => e
puts e
end
puts
end
```
Output:
```Given points:
[0.1234, 0.9876],
[0.8765, 0.2345]
You can construct the following circles:
#<struct Circle x=1.8631118016581891, y=1.974211801658189, r=2.0>
#<struct Circle x=-0.8632118016581893, y=-0.752111801658189, r=2.0>

Given points:
[0.0, 2.0],
[0.0, 0.0]
You can construct the following circles:
#<struct Circle x=0.0, y=1.0, r=1.0>

Given points:
[0.1234, 0.9876],
[0.1234, 0.9876]
Infinite number of circles, points coincide.

Given points:
[0.1234, 0.9876],
[0.8765, 0.2345]
Distance of points > diameter.

Given points:
[0.1234, 0.9876],
[0.1234, 0.9876]
You can construct the following circles:
#<struct Circle x=0.1234, y=0.9876, r=0.0>
```

## Run BASIC

```html "<TABLE border=1>"
html "<tr bgcolor=wheat align=center><td>No.</td><td>x1</td><td>y1</td><td>x2</td><td>y2</td><td>r</td><td>cir x1</td><td>cir y1</td><td>cir x2</td><td>cir y2</td></tr>"
for i = 1 to 5
html "<tr align=right><td>";i;"</td><td>";x1;"</td><td>";y1;"</td><td>";x2;"</td><td>";y2;"</td><td>";r;"</td>"
gosub [twoCircles]
next
html "</table>"
end

'p1                p2           r
data 0.1234, 0.9876,    0.8765, 0.2345,    2.0
data 0.0000, 2.0000,    0.0000, 0.0000,    1.0
data 0.1234, 0.9876,    0.1234, 0.9876,    2.0
data 0.1234, 0.9876,    0.8765, 0.2345,    0.5
data 0.1234, 0.9876,    0.1234, 0.9876,    0.0

[twoCircles]

if x1=x2 and y1=y2 then '2.If the points are coincident
if r=0 then ' unless r==0.0
html "<td colspan=4 align=left>It will be a single point (";x1;",";y1;") of radius 0</td></tr>"
RETURN
else
html "<td colspan=4 align=left>There are any number of circles via single point (";x1;",";y1;") of radius ";r;"</td></tr>"
RETURN
end if
end if
r2 = sqr((x1-x2)^2+(y1-y2)^2)/2			'half distance between points
if r<r2 then
html "<td colspan=4 align=left>Points are too far apart (";2*r2;") - there are no circles of radius ";r
RETURN
end if

'else, calculate two centers
cx=(x1+x2)/2 					'middle point
cy=(y1+y2)/2
'should move from middle point along perpendicular by dd2
dd2=sqr(r^2-r2^2)					'perpendicular distance
dx1=x2-cx   					'vector to middle point
dy1=y2-cy
dx = 0-dy1/r2*dd2   				'perpendicular:
dy = dx1/r2*dd2     				'rotate and scale
html "<td>";cx+dy;"</td><td>";cy+dx;"</td>"   	'two points, with (+)
html "<td>";cx-dy;"</td><td>";cy-dx;"</td></TR>" 	'and (-)
RETURN```
Output:
 No. x1 y1 x2 y2 r cir x1 cir y1 cir x2 cir y2 1 0.1234 0.9876 0.8765 0.2345 2.0 1.8631118 1.9742118 -0.863211802 -0.752111802 2 0.0d 2.0 0.0d 0.0d 1.0 0.0d 1.0 0.0d 1.0 3 0.1234 0.9876 0.1234 0.9876 2.0 There are any number of circles via single point (0.1234,0.9876) of radius 2.0 4 0.1234 0.9876 0.8765 0.2345 0.5 Points are too far apart (1.06504423) - there are no circles of radius 0.5 5 0.1234 0.9876 0.1234 0.9876 0.0d It will be a single point (0.1234,0.9876) of radius 0

## Rust

Translation of: C
```use std::fmt;

#[derive(Clone,Copy)]
struct Point {
x: f64,
y: f64
}

fn distance (p1: Point, p2: Point) -> f64 {
((p1.x - p2.x).powi(2) + (p1.y - p2.y).powi(2)).sqrt()
}

impl fmt::Display for Point {
fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
write!(f, "({:.4}, {:.4})", self.x, self.y)
}
}

fn describe_circle(p1: Point, p2: Point, r: f64) {
let sep = distance(p1, p2);

if sep == 0. {
if r == 0. {
println!("No circles can be drawn through {}", p1);
} else {
println!("Infinitely many circles can be drawn through {}", p1);
}
} else if sep == 2.0 * r {
println!("Given points are opposite ends of a diameter of the circle with center ({:.4},{:.4}) and r {:.4}",
(p1.x+p2.x) / 2.0, (p1.y+p2.y) / 2.0, r);
} else if sep > 2.0 * r {
println!("Given points are farther away from each other than a diameter of a circle with r {:.4}", r);
} else {
let mirror_dist = (r.powi(2) - (sep / 2.0).powi(2)).sqrt();

println!("Two circles are possible.");
println!("Circle C1 with center ({:.4}, {:.4}), r {:.4} and Circle C2 with center ({:.4}, {:.4}), r {:.4}",
((p1.x + p2.x) / 2.0) + mirror_dist * (p1.y-p2.y)/sep, (p1.y+p2.y) / 2.0 + mirror_dist*(p2.x-p1.x)/sep,
r,
(p1.x+p2.x) / 2.0 - mirror_dist*(p1.y-p2.y)/sep, (p1.y+p2.y) / 2.0 - mirror_dist*(p2.x-p1.x)/sep, r);
}
}

fn main() {
let points: Vec<(Point, Point)> = vec![
(Point { x: 0.1234, y: 0.9876 }, Point { x: 0.8765, y: 0.2345 }),
(Point { x: 0.0000, y: 2.0000 }, Point { x: 0.0000, y: 0.0000 }),
(Point { x: 0.1234, y: 0.9876 }, Point { x: 0.1234, y: 0.9876 }),
(Point { x: 0.1234, y: 0.9876 }, Point { x: 0.8765, y: 0.2345 }),
(Point { x: 0.1234, y: 0.9876 }, Point { x: 0.1234, y: 0.9876 })
];
let radii: Vec<f64> = vec![2.0, 1.0, 2.0, 0.5, 0.0];

for (p, r) in points.into_iter().zip(radii.into_iter()) {
println!("\nPoints: ({}, {}), Radius: {:.4}", p.0, p.1, r);
describe_circle(p.0, p.1, r);
}
}
```
Output:
```Points: ((0.1234, 0.9876), (0.8765, 0.2345)), Radius: 2.0000
Two circles are possible.
Circle C1 with center (1.8631, 1.9742), r 2.0000 and Circle C2 with center (-0.8632, -0.7521), r 2.0000

Points: ((0.0000, 2.0000), (0.0000, 0.0000)), Radius: 1.0000
Given points are opposite ends of a diameter of the circle with center (0.0000,1.0000) and r 1.0000

Points: ((0.1234, 0.9876), (0.1234, 0.9876)), Radius: 2.0000
Infinitely many circles can be drawn through (0.1234, 0.9876)

Points: ((0.1234, 0.9876), (0.8765, 0.2345)), Radius: 0.5000
Given points are farther away from each other than a diameter of a circle with r 0.5000

Points: ((0.1234, 0.9876), (0.1234, 0.9876)), Radius: 0.0000
No circles can be drawn through (0.1234, 0.9876)```

## Scala

```import org.scalatest.FunSuite
import math._

case class V2(x: Double, y: Double) {
val distance = hypot(x, y)
def /(other: V2) = V2((x+other.x) / 2.0, (y+other.y) / 2.0)
def -(other: V2) = V2(x-other.x,y-other.y)
override def equals(other: Any) = other match {
case p: V2 => abs(x-p.x) <  0.0001 && abs(y-p.y) <  0.0001
case _ => false
}
override def toString = f"(\$x%.4f, \$y%.4f)"
}

case class Circle(center: V2, radius: Double)

class PointTest extends FunSuite {
println("       p1               p2         r    result")
Seq(
(V2(0.1234, 0.9876), V2(0.8765, 0.2345), 2.0, Seq(Circle(V2(1.8631, 1.9742), 2.0), Circle(V2(-0.8632, -0.7521), 2.0))),
(V2(0.0000, 2.0000), V2(0.0000, 0.0000), 1.0, Seq(Circle(V2(0.0, 1.0), 1.0))),
(V2(0.1234, 0.9876), V2(0.1234, 0.9876), 2.0, "coincident points yields infinite circles"),
(V2(0.1234, 0.9876), V2(0.8765, 0.2345), 0.5, "radius is less then the distance between points"),
(V2(0.1234, 0.9876), V2(0.1234, 0.9876), 0.0, "radius of zero yields no circles")
) foreach { v =>
print(s"\${v._1} \${v._2}  \${v._3}: ")
circles(v._1, v._2, v._3) match {
case Right(list) => println(list mkString ",")
assert(list === v._4)
case Left(error) => println(error)
assert(error === v._4)
}
}

def circles(p1: V2, p2: V2, radius: Double) = if (radius == 0.0) {
Left("radius of zero yields no circles")
} else if (p1 == p2) {
Left("coincident points yields infinite circles")
} else if (radius * 2 < (p1-p2).distance) {
Left("radius is less then the distance between points")
} else {
} ensuring { result =>
result.isLeft || result.right.get.nonEmpty
}

def circlesThruPoints(p1: V2, p2: V2, radius: Double): Seq[Circle] = {
val diff = p2 - p1
val d = pow(pow(radius, 2) - pow(diff.distance / 2, 2), 0.5)
val mid = p1 / p2
Seq(
Circle(V2(mid.x - d * diff.y / diff.distance, mid.y + d * diff.x / diff.distance), abs(radius)),
Circle(V2(mid.x + d * diff.y / diff.distance, mid.y - d * diff.x / diff.distance), abs(radius))).distinct
}
}
```
Output:
```       p1               p2         r    result
(0.1234, 0.9876) (0.8765, 0.2345)  2.0: Circle((1.8631, 1.9742),2.0),Circle((-0.8632, -0.7521),2.0)
(0.0000, 2.0000) (0.0000, 0.0000)  1.0: Circle((0.0000, 1.0000),1.0)
(0.1234, 0.9876) (0.1234, 0.9876)  2.0: coincident points yields infinite circles
(0.1234, 0.9876) (0.8765, 0.2345)  0.5: radius is less then the distance between points
(0.1234, 0.9876) (0.1234, 0.9876)  0.0: radius of zero yields no circlesEmpty test suite.```

## Scheme

```(import (scheme base)
(scheme inexact)
(scheme write))

;; c1 and c2 are pairs (x y), r a positive radius
(define (find-circles c1 c2 r)
(define x-coord car) ; for easier to read coordinate extraction from list
(define (approx= a b) (< (- a b) 0.000001)) ; equal within tolerance
(define (avg a b) (/ (+ a b) 2))
(define (distance pt1 pt2)
(sqrt (+ (square (- (x-coord pt1) (x-coord pt2)))
(square (- (y-coord pt1) (y-coord pt2))))))
(define (equal-points? pt1 pt2)
(and (approx= (x-coord pt1) (x-coord pt2))
(approx= (y-coord pt1) (y-coord pt2))))
(define (delete-duplicate pts) ; assume no more than two points in list
(if (and (= 2 (length pts))
(list (car pts)) ; keep the first only
pts))
;
(let ((d (distance c1 c2)))
(cond ((equal-points? c1 c2) ; coincident points
(if (> r 0)
'infinite   ; r > 0
(list c1))) ; else r = 0
((< (* 2 r) d)
'()) ; circle cannot reach both points, as too far apart
((approx= r 0.0) ; r = 0, no circles, as points differ
'())
(else ; find up to two circles meeting c1 and c2
(let* ((mid-pt (list (avg (x-coord c1) (x-coord c2))
(avg (y-coord c1) (y-coord c2))))
(offset (sqrt (- (square r)
(square (* 0.5 d)))))
(delta-cx (/ (- (x-coord c1) (x-coord c2)) d))
(delta-cy (/ (- (y-coord c1) (y-coord c2)) d)))
(delete-duplicate
(list (list (- (x-coord mid-pt) (* offset delta-cx))
(+ (y-coord mid-pt) (* offset delta-cy)))
(list (+ (x-coord mid-pt) (* offset delta-cx))
(- (y-coord mid-pt) (* offset delta-cy))))))))))

;; work through the input examples, outputting results
(for-each
(lambda (c1 c2 r)
(let ((result (find-circles c1 c2 r)))
(display "p1: ") (display c1)
(display " p2: ") (display c2)
(display " r: ") (display (number->string r))
(display " => ")
(cond ((eq? result 'infinite)
(display "Infinite number of circles"))
((null? result)
(display "No circles"))
(else
(display result)))
(newline)))
'((0.1234 0.9876) (0.0000 2.0000) (0.1234 0.9876) (0.1234 0.9876) (0.1234 0.9876))
'((0.8765 0.2345) (0.0000 0.0000) (0.1234 0.9876) (0.8765 0.2345) (0.1234 0.9876))
'(2.0 1.0 2.0 0.5 0.0))
```
Output:
```p1: (0.1234 0.9876) p2: (0.8765 0.2345) r: 2.0 => ((1.86311180165819 1.97421180165819) (-0.863211801658189 -0.752111801658189))
p1: (0.0 2.0) p2: (0.0 0.0) r: 1.0 => ((0.0 1.0))
p1: (0.1234 0.9876) p2: (0.1234 0.9876) r: 2.0 => Infinite number of circles
p1: (0.1234 0.9876) p2: (0.8765 0.2345) r: 0.5 => No circles
p1: (0.1234 0.9876) p2: (0.1234 0.9876) r: 0.0 => ((0.1234 0.9876))
```

## Seed7

```\$ include "seed7_05.s7i";
include "float.s7i";
include "math.s7i";

const type: point is new struct
var float: x is 0.0;
var float: y is 0.0;
end struct;

const func point: point (in float: x, in float: y) is func
result
var point: aPoint is point.value;
begin
aPoint.x := x;
aPoint.y := y;
end func;

const func float: distance (in point: p1, in point: p2) is
return sqrt((p1.x - p2.x) ** 2 + (p1.y - p2.y) ** 2);

const proc: findCircles (in point: p1, in point: p2, in float: radius) is func
local
var float: separation is 0.0;
var float: mirrorDistance is 0.0;
begin
separation := distance(p1, p2);
if separation = 0.0 then
write("Radius of zero. No circles can be drawn through (");
else
write("Infinitely many circles can be drawn through (");
end if;
writeln(p1.x digits 4 <& ", " <& p1.y digits 4 <& ")");
elsif separation = 2.0 * radius then
writeln("Given points are opposite ends of a diameter of the circle with center (" <&
(p1.x + p2.x) / 2.0 digits 4 <& ", " <& (p1.y + p2.y) / 2.0 digits 4 <& ") and radius " <&
elsif separation > 2.0 * radius then
writeln("Given points are farther away from each other than a diameter of a circle with radius " <&
else
mirrorDistance := sqrt(radius ** 2 - (separation / 2.0) ** 2);
writeln("Two circles are possible.");
writeln("Circle C1 with center (" <&
(p1.x + p2.x) / 2.0 + mirrorDistance*(p1.y - p2.y) / separation digits 4 <& ", " <&
(p1.y + p2.y) / 2.0 + mirrorDistance*(p2.x - p1.x) / separation digits 4 <& "), radius " <&
writeln("Circle C2 with center (" <&
(p1.x + p2.x) / 2.0 - mirrorDistance*(p1.y - p2.y) / separation digits 4 <& ", " <&
(p1.y + p2.y) / 2.0 - mirrorDistance*(p2.x - p1.x) / separation digits 4 <& "), radius " <&
end if;
end func;

const proc: main is func
local
const array array float: cases is [] (
[] (0.1234, 0.9876, 0.8765, 0.2345, 2.0),
[] (0.0000, 2.0000, 0.0000, 0.0000, 1.0),
[] (0.1234, 0.9876, 0.1234, 0.9876, 2.0),
[] (0.1234, 0.9876, 0.8765, 0.2345, 0.5),
[] (0.1234, 0.9876, 0.1234, 0.9876, 0.0));
var integer: index is 0;
begin
for index range 1 to 5 do
writeln("Case " <& index <& ":");
findCircles(point(cases[index][1], cases[index][2]),
point(cases[index][3], cases[index][4]), cases[index][5]);
end for;
end func;```
Output:
```Case 1:
Two circles are possible.
Circle C1 with center (1.8631, 1.9742), radius 2.0000
Circle C2 with center (-0.8632, -0.7521), radius 2.0000
Case 2:
Given points are opposite ends of a diameter of the circle with center (0.0000, 1.0000) and radius 1.0000
Case 3:
Infinitely many circles can be drawn through (0.1234, 0.9876)
Case 4:
Given points are farther away from each other than a diameter of a circle with radius 0.5000
Case 5:
Radius of zero. No circles can be drawn through (0.1234, 0.9876)
```

## Sidef

Translation of: Raku
```func circles(a, b, r) {

if (a == b) {
if (r == 0) {
return ['Degenerate point']
}
else {
return ['Infinitely many share a point']
}
}

var h = (b-a)/2

if (r**2 < h.norm) {
return ['Too far apart']
}

var l = sqrt(r**2 - h.norm)

[1i, -1i].map {|i|
a + h + (l*i*h / h.abs) -> round(-16)
}
}

var input = [
[0.1234 + 0.9876i,  0.8765 + 0.2345i, 2.0],
[0.0000 + 2.0000i,  0.0000 + 0.0000i, 1.0],
[0.1234 + 0.9876i,  0.1234 + 0.9876i, 2.0],
[0.1234 + 0.9876i,  0.8765 + 0.2345i, 0.5],
[0.1234 + 0.9876i,  0.1234 + 0.9876i, 0.0],
]

input.each {|a|
say (a.join(', '), ': ', circles(a...).join(' and '))
}
```
Output:
```0.1234+0.9876i, 0.8765+0.2345i, 2: 1.8631118016581891+1.9742118016581891i and -0.8632118016581891-0.7521118016581891i
2i, 0, 1: i and i
0.1234+0.9876i, 0.1234+0.9876i, 2: Infinitely many share a point
0.1234+0.9876i, 0.8765+0.2345i, 0.5: Too far apart
0.1234+0.9876i, 0.1234+0.9876i, 0: Degenerate point
```

## Stata

Each circle center is the image of B by the composition of a rotation and homothecy centered at A. It's how the centers are computed in this implementation. The coordinates are returned as the columns of a 2x2 matrix. When the solution is not unique or does not exist, this matrix contains only missing values.

```real matrix centers(real colvector a, real colvector b, real scalar r) {
real matrix rot
real scalar d, u, v
d = norm(b-a)
if (r == 0 | d == 0) {
if (r == 0 & d == 0) {
return((a,a))
} else {
return(J(2, 2, .))
}
} else if (d <= 2*r) {
u = d/(2*r)
v = sqrt(1-u^2)
rot = u,-v\v,u
return((rot*(b-a),rot'*(b-a))*r/d:+a)
} else {
return(J(2, 2, .))
}
}
```

Examples:

```:a=0.1234\0.9876
:b=0.8765\0.2345
: centers(a,b,2)
1              2
+-------------------------------+
1 |   1.863111802   -.8632118017  |
2 |   1.974211802   -.7521118017  |
+-------------------------------+

: centers((0\2),(0\0),1)
1   2
+---------+
1 |  0   0  |
2 |  1   1  |
+---------+

: centers(a,a,2)
[symmetric]
1   2
+---------+
1 |  .      |
2 |  .   .  |
+---------+

: centers(a,b,0.5)
[symmetric]
1   2
+---------+
1 |  .      |
2 |  .   .  |
+---------+

: centers(a,a,0)
1       2
+-----------------+
1 |  .1234   .1234  |
2 |  .9876   .9876  |
+-----------------+
```

## Swift

Translation of: F#
```import Foundation

struct Point: Equatable {
var x: Double
var y: Double
}

struct Circle {
var center: Point

static func circleBetween(
_ p1: Point,
_ p2: Point,
) -> (Circle, Circle?)? {
func applyPoint(_ p1: Point, _ p2: Point, op: (Double, Double) -> Double) -> Point {
return Point(x: op(p1.x, p2.x), y: op(p1.y, p2.y))
}

func mul2(_ p: Point, mul: Double) -> Point {
return Point(x: p.x * mul, y: p.y * mul)
}

func div2(_ p: Point, div: Double) -> Point {
return Point(x: p.x / div, y: p.y / div)
}

func norm(_ p: Point) -> Point {
return div2(p, div: (p.x * p.x + p.y * p.y).squareRoot())
}

guard radius != 0, p1 != p2 else {
return nil
}

let diameter = 2 * radius
let pq = applyPoint(p1, p2, op: -)
let magPQ = (pq.x * pq.x + pq.y * pq.y).squareRoot()

guard diameter >= magPQ else {
return nil
}

let midpoint = div2(applyPoint(p1, p2, op: +), div: 2)
let halfPQ = magPQ / 2
let midC = mul2(norm(Point(x: -pq.y, y: pq.x)), mul: magMidC)
let center1 = applyPoint(midpoint, midC, op: +)
let center2 = applyPoint(midpoint, midC, op: -)

if center1 == center2 {
} else {
}
}
}

let testCases = [
(Point(x: 0.1234, y: 0.9876), Point(x: 0.8765, y: 0.2345), 2.0),
(Point(x: 0.0000, y: 2.0000), Point(x: 0.0000, y: 0.0000), 1.0),
(Point(x: 0.1234, y: 0.9876), Point(x: 0.1234, y: 0.9876), 2.0),
(Point(x: 0.1234, y: 0.9876), Point(x: 0.8765, y: 0.2345), 0.5),
(Point(x: 0.1234, y: 0.9876), Point(x: 0.1234, y: 0.9876), 0.0)
]

for testCase in testCases {
switch Circle.circleBetween(testCase.0, testCase.1, withRadius: testCase.2) {
case nil:
print("No ans")
case (let circle1, nil)?:
print("One ans: \(circle1)")
case (let circle1, let circle2?)?:
print("Two ans: \(circle1) \(circle2)")
}
}
```
Output:
```Two ans: Circle(center: Point(x: -0.8632118016581896, y: -0.7521118016581892), radius: 2.0) Circle(center: Point(x: 1.8631118016581893, y: 1.974211801658189), radius: 2.0)
One ans: Circle(center: Point(x: 0.0, y: 1.0), radius: 1.0)
No ans
No ans
No ans```

## Tcl

Translation of: Python
```proc findCircles {p1 p2 r} {
lassign \$p1 x1 y1
lassign \$p2 x2 y2
# Special case: coincident & zero size
if {\$x1 == \$x2 && \$y1 == \$y2 && \$r == 0.0} {
return [list [list \$x1 \$y1 0.0]]
}
if {\$r <= 0.0} {
error "radius must be positive for sane results"
}
if {\$x1 == \$x2 && \$y1 == \$y2} {
error "no sane solution: points are coincident"
}

# Calculate distance apart and separation vector
set dx [expr {\$x2 - \$x1}]
set dy [expr {\$y2 - \$y1}]
set q [expr {hypot(\$dx, \$dy)}]
if {\$q > 2*\$r} {
error "no solution: points are further apart than required diameter"
}

# Calculate midpoint
set x3 [expr {(\$x1+\$x2)/2.0}]
set y3 [expr {(\$y1+\$y2)/2.0}]
# Fractional distance along the mirror line
set f [expr {(\$r**2 - (\$q/2.0)**2)**0.5 / \$q}]
set c1 [list [expr {\$x3 - \$f*\$dy}] [expr {\$y3 + \$f*\$dx}] \$r]
set c2 [list [expr {\$x3 + \$f*\$dy}] [expr {\$y3 - \$f*\$dx}] \$r]
return [list \$c1 \$c2]
}
```
Demo:
```foreach {p1 p2 r} {
{0.1234 0.9876} {0.8765 0.2345} 2.0
{0.0000 2.0000} {0.0000 0.0000} 1.0
{0.1234 0.9876} {0.1234 0.9876} 2.0
{0.1234 0.9876} {0.8765 0.2345} 0.5
{0.1234 0.9876} {0.1234 0.9876} 0.0
} {
puts "p1:([join \$p1 {, }]) p2:([join \$p2 {, }]) r:\$r =>"
if {[catch {
foreach c [findCircles \$p1 \$p2 \$r] {
puts "\tCircle:([join \$c {, }])"
}
} msg]} {
puts "\tERROR: \$msg"
}
}
```
Output:
```p1:(0.1234, 0.9876) p2:(0.8765, 0.2345) r:2.0 =>
Circle:(1.863111801658189, 1.974211801658189, 2.0)
Circle:(-0.8632118016581891, -0.752111801658189, 2.0)
p1:(0.0000, 2.0000) p2:(0.0000, 0.0000) r:1.0 =>
Circle:(0.0, 1.0, 1.0)
Circle:(0.0, 1.0, 1.0)
p1:(0.1234, 0.9876) p2:(0.1234, 0.9876) r:2.0 =>
ERROR: no sane solution: points are coincident
p1:(0.1234, 0.9876) p2:(0.8765, 0.2345) r:0.5 =>
ERROR: no solution: points are further apart than required diameter
p1:(0.1234, 0.9876) p2:(0.1234, 0.9876) r:0.0 =>
Circle:(0.1234, 0.9876, 0.0)
```

## VBA

Translation of: Phix
```Public Sub circles()
tests = [{0.1234, 0.9876, 0.8765, 0.2345, 2.0; 0.0000, 2.0000, 0.0000, 0.0000, 1.0; 0.1234, 0.9876, 0.1234, 0.9876, 2.0; 0.1234, 0.9876, 0.8765, 0.2345, 0.5; 0.1234, 0.9876, 0.1234, 0.9876, 0.0}]
For i = 1 To UBound(tests)
x1 = tests(i, 1)
y1 = tests(i, 2)
x2 = tests(i, 3)
y2 = tests(i, 4)
R = tests(i, 5)
xd = x2 - x1
yd = y1 - y2
s2 = xd * xd + yd * yd
sep = Sqr(s2)
xh = (x1 + x2) / 2
yh = (y1 + y2) / 2
Dim txt As String
If sep = 0 Then
txt = "same points/" & IIf(R = 0, "radius is zero", "infinite solutions")
Else
If sep = 2 * R Then
txt = "opposite ends of diameter with centre " & xh & ", " & yh & "."
Else
If sep > 2 * R Then
txt = "too far apart " & sep & " > " & 2 * R
Else
md = Sqr(R * R - s2 / 4)
xs = md * xd / sep
ys = md * yd / sep
txt = "{" & Format(xh + ys, "0.0000") & ", " & Format(yh + xs, "0.0000") & _
"} and {" & Format(xh - ys, "0.0000") & ", " & Format(yh - xs, "0.0000") & "}"
End If
End If
End If
Debug.Print "points " & "{" & x1 & ", " & y1 & "}" & ", " & "{" & x2 & ", " & y2 & "}" & " with radius " & R & " ==> " & txt
Next i
End Sub```
Output:
```points {0,1234, 0,9876}, {0,8765, 0,2345} with radius 2 ==> {1,8631, 1,9742} and {-0,8632, -0,7521}
points {0, 2}, {0, 0} with radius 1 ==> opposite ends of diameter with centre 0, 1.
points {0,1234, 0,9876}, {0,1234, 0,9876} with radius 2 ==> same points/infinite solutions
points {0,1234, 0,9876}, {0,8765, 0,2345} with radius 0,5 ==> too far apart 1,06504423382318 > 1
points {0,1234, 0,9876}, {0,1234, 0,9876} with radius 0 ==> same points/radius is zero```

## Visual Basic .NET

Translation of: C#
```Public Class CirclesOfGivenRadiusThroughTwoPoints
Public Shared Sub Main()
For Each valu In New Double()() {
New Double() {0.1234, 0.9876, 0.8765, 0.2345, 2},
New Double() {0.0, 2.0, 0.0, 0.0, 1},
New Double() {0.1234, 0.9876, 0.1234, 0.9876, 2},
New Double() {0.1234, 0.9876, 0.8765, 0.2345, 0.5},
New Double() {0.1234, 0.9876, 0.1234, 0.9876, 0},
New Double() {0.1234, 0.9876, 0.2345, 0.8765, 0}}
Dim p = New Point(valu(0), valu(1)), q = New Point(valu(2), valu(3))
Console.WriteLine(\$"Points {p} and {q} with radius {valu(4)}:")
Try
Console.WriteLine(vbTab & String.Join(" and ", FindCircles(p, q, valu(4))))
Catch ex As Exception
Console.WriteLine(vbTab & ex.Message)
End Try
Next
End Sub

Private Shared Function FindCircles(ByVal p As Point, ByVal q As Point, ByVal rad As Double) As Point()
If rad = 0 Then Throw New InvalidOperationException(If(p = q,
String.Format("{0} (degenerate circle)", {p}), "No circles."))
If p = q Then Throw New InvalidOperationException("Infinite number of circles.")
Dim dist As Double = Point.Distance(p, q), sqDist As Double = dist * dist,
If sqDist > sqDiam Then Throw New InvalidOperationException(
String.Format("Points are too far apart (by {0}).", sqDist - sqDiam))
Dim midPoint As Point = New Point((p.X + q.X) / 2, (p.Y + q.Y) / 2)
If sqDist = sqDiam Then Return {midPoint}
Dim d As Double = Math.Sqrt(rad * rad - sqDist / 4),
a As Double = d * (q.X - p.X) / dist, b As Double = d * (q.Y - p.Y) / dist
Return {New Point(midPoint.X - b, midPoint.Y + a), New Point(midPoint.X + b, midPoint.Y - a)}
End Function

Public Structure Point
Public ReadOnly Property X As Double
Public ReadOnly Property Y As Double

Public Sub New(ByVal ix As Double, ByVal iy As Double)
Me.New() : X = ix : Y = iy
End Sub

Public Shared Operator =(ByVal p As Point, ByVal q As Point) As Boolean
Return p.X = q.X AndAlso p.Y = q.Y
End Operator

Public Shared Operator <>(ByVal p As Point, ByVal q As Point) As Boolean
Return p.X <> q.X OrElse p.Y <> q.Y
End Operator

Public Shared Function SquaredDistance(ByVal p As Point, ByVal q As Point) As Double
Dim dx As Double = q.X - p.X, dy As Double = q.Y - p.Y
Return dx * dx + dy * dy
End Function

Public Shared Function Distance(ByVal p As Point, ByVal q As Point) As Double
Return Math.Sqrt(SquaredDistance(p, q))
End Function

Public Overrides Function ToString() As String
Return \$"({X}, {Y})"
End Function
End Structure
End Class
```
Output:
```Points (0.1234, 0.9876) and (0.8765, 0.2345) with radius 2:
(1.86311180165819, 1.97421180165819) and (-0.86321180165819, -0.752111801658189)
Points (0, 2) and (0, 0) with radius 1:
(0, 1)
Points (0.1234, 0.9876) and (0.1234, 0.9876) with radius 2:
Infinite number of circles.
Points (0.1234, 0.9876) and (0.8765, 0.2345) with radius 0.5:
Points are too far apart (by 0.13431922).
Points (0.1234, 0.9876) and (0.1234, 0.9876) with radius 0:
(0.1234, 0.9876) (degenerate circle)
Points (0.1234, 0.9876) and (0.2345, 0.8765) with radius 0:
No circles.```

## Visual FoxPro

Translation of BASIC.

```LOCAL p1 As point, p2 As point, rr As Double
CLOSE DATABASES ALL
SET FIXED ON
SET DECIMALS TO 4
CLEAR
CREATE CURSOR circles (xc1 B(4), yc1 B(4), xc2 B(4), yc2 B(4), rad B(4))
INSERT INTO circles VALUES (0.1234, 0.9876, 0.8765, 0.2345, 2.0)
INSERT INTO circles VALUES (0.0000, 2.0000, 0.0000, 0.0000, 1.0)
INSERT INTO circles VALUES (0.1234, 0.9876, 0.1234, 0.9876, 2.0)
INSERT INTO circles VALUES (0.1234, 0.9876, 0.8765, 0.2345, 0.5)
INSERT INTO circles VALUES (0.1234, 0.9876, 0.1234, 0.9876, 0.0)
GO TOP

p1 = NEWOBJECT("point")
p2 = NEWOBJECT("point")
SCAN
p1.SetPoints(xc1, yc1)
p2.SetPoints(xc2, yc2)
GetCircles(p1, p2, rr)
?
ENDSCAN

SET DECIMALS TO
SET FIXED OFF

PROCEDURE GetCircles(op1 As point, op2 As point, r As Double)
LOCAL ctr As point, half As point, lenhalf As Double, dist As Double, rot As point, c As String
ctr = NEWOBJECT("point")
half = NEWOBJECT("point")
ctr.SetPoints((op1.xc + op2.xc)/2, (op1.yc + op2.yc)/2)
half.SetPoints(op1.xc - ctr.xc, op1.yc - ctr.yc)
lenhalf = half.nLength
PrintPoints(op1, op2, r)
IF r < lenhalf
? "Cannot solve for these parameters."
RETURN
ENDIF
IF lenhalf = 0
? "Points are coincident."
RETURN
ENDIF
dist = SQRT(r^2 - lenhalf^2)/lenhalf
rot = NEWOBJECT("point")
rot.SetPoints(-dist*(op1.yc - ctr.yc) + ctr.xc, dist*(op1.xc - ctr.xc) + ctr.yc)
TEXT TO c TEXTMERGE NOSHOW PRETEXT 3
Circle 1 (<<rot.xc>>, <<rot.yc>>)
ENDTEXT
? c
rot.SetPoints(-(rot.xc - ctr.xc) + ctr.xc, -((rot.yc - ctr.yc)) + ctr.yc)
TEXT TO c TEXTMERGE NOSHOW PRETEXT 3
Circle 2 (<<rot.xc>>, <<rot.yc>>)
ENDTEXT
? c
ENDPROC

PROCEDURE PrintPoints(op1 As point, op2 As point, r As Double)
LOCAL lcTxt As String
TEXT TO lcTxt TEXTMERGE NOSHOW PRETEXT 3
ENDTEXT
? lcTxt
ENDPROC

DEFINE CLASS point As Custom
xc = 0
yc = 0
nLength = 0

PROCEDURE Init
DODEFAULT()
ENDPROC

PROCEDURE SetPoints(tnx As Double, tny As Double)
THIS.xc = tnx
THIS.yc = tny
THIS.nLength = THIS.GetLength()
ENDPROC

FUNCTION GetLength()
RETURN SQRT(THIS.xc*THIS.xc + THIS.yc*THIS.yc)
ENDFUNC

ENDDEFINE
```
Output:
```
Circle 1 (-0.8632, -0.7521)
Circle 1 (-0.8632, -0.7521)
Circle 2 (1.8631, 1.9742)
Circle 2 (1.8631, 1.9742)

Circle 1 (0.0000, 1.0000)
Circle 1 (0.0000, 1.0000)
Circle 2 (0.0000, 1.0000)
Circle 2 (0.0000, 1.0000)

Points are coincident.

Cannot solve for these parameters.

Points are coincident.
```

## V (Vlang)

Translation of: Go
```import math

const (
two  = "two circles."
r0   = "R==0.0 does not describe circles."
co   = "coincident points describe an infinite number of circles."
cor0 = "coincident points with r==0.0 describe a degenerate circle."
diam = "Points form a diameter and describe only a single circle."
far  = "Points too far apart to form circles."
)

struct Point {
x f64
y f64
}

fn circles(p1 Point, p2 Point, r f64) (Point, Point, string) {
mut case := ''
c1, c2 := p1, p2
if p1 == p2 {
if r == 0 {
return p1, p1, cor0
}
case = co
return c1, c2, case
}
if r == 0 {
return p1, p2, r0
}
dx := p2.x - p1.x
dy := p2.y - p1.y
q := math.hypot(dx, dy)
if q > 2*r {
case = far
return c1, c2, case
}
m := Point{(p1.x + p2.x) / 2, (p1.y + p2.y) / 2}
if q == 2*r {
return m, m, diam
}
d := math.sqrt(r*r - q*q/4)
ox := d * dx / q
oy := d * dy / q
return Point{m.x - oy, m.y + ox}, Point{m.x + oy, m.y - ox}, two
}

struct Cir {
p1 Point
p2 Point
r f64
}
const td = [
Cir{Point{0.1234, 0.9876}, Point{0.8765, 0.2345}, 2.0},
Cir{Point{0.0000, 2.0000}, Point{0.0000, 0.0000}, 1.0},
Cir{Point{0.1234, 0.9876}, Point{0.1234, 0.9876}, 2.0},
Cir{Point{0.1234, 0.9876}, Point{0.8765, 0.2345}, 0.5},
Cir{Point{0.1234, 0.9876}, Point{0.1234, 0.9876}, 0.0},
]

fn main() {
for tc in td {
println("p1:  \$tc.p1")
println("p2:  \$tc.p2")
println("r:  \$tc.r")
c1, c2, case := circles(tc.p1, tc.p2, tc.r)
println("   \$case")
match case {
cor0, diam{
println("   Center:  \$c1")
}
two {
println("   Center 1:  \$c1")
println("   Center 2:  \$c2")
}
else{}
}
println('')
}
}```
Output:
```p1:  Point{
x: 0.1234
y: 0.9876
}
p2:  Point{
x: 0.8765
y: 0.2345
}
r:  2
two circles.
Center 1:  Point{
x: 1.863111801658189
y: 1.9742118016581887
}
Center 2:  Point{
x: -0.8632118016581891
y: -0.7521118016581888
}

p1:  Point{
x: 0
y: 2
}
p2:  Point{
x: 0
y: 0
}
r:  1
Points form a diameter and describe only a single circle.
Center:  Point{
x: 0
y: 1
}

p1:  Point{
x: 0.1234
y: 0.9876
}
p2:  Point{
x: 0.1234
y: 0.9876
}
r:  2
coincident points describe an infinite number of circles.

p1:  Point{
x: 0.1234
y: 0.9876
}
p2:  Point{
x: 0.8765
y: 0.2345
}
r:  0.5
Points too far apart to form circles.

p1:  Point{
x: 0.1234
y: 0.9876
}
p2:  Point{
x: 0.1234
y: 0.9876
}
r:  0
coincident points with r==0.0 describe a degenerate circle.
Center:  Point{
x: 0.1234
y: 0.9876
}
```

## Wren

Translation of: Go
Library: Wren-math
```import "./math" for Math

var Two  = "Two circles."
var R0   = "R == 0 does not describe circles."
var Co   = "Coincident points describe an infinite number of circles."
var CoR0 = "Coincident points with r == 0 describe a degenerate circle."
var Diam = "Points form a diameter and describe only a single circle."
var Far  = "Points too far apart to form circles."

class Point {
construct new(x, y) {
_x = x
_y = y
}

x { _x }
y { _y }
==(p) { _x == p.x && _y == p.y }

toString { "(%(_x), %(_y))" }
}

var circles = Fn.new { |p1, p2, r|
var c1 = Point.new(0, 0)
var c2 = Point.new(0, 0)
if (p1 == p2) {
if (r == 0) return [p1, p1, CoR0]
return [c1, c2, Co]
}
if (r == 0) return [p1, p2, R0]
var dx = p2.x - p1.x
var dy = p2.y - p1.y
var q = Math.hypot(dx, dy)
if (q > 2*r) return [c1, c2, Far]
var m = Point.new((p1.x + p2.x)/2, (p1.y + p2.y)/2)
if (q == 2*r) return [m, m, Diam]
var d = (r*r - q*q/4).sqrt
var ox = d * dx / q
var oy = d * dy / q
return [Point.new(m.x - oy, m.y + ox), Point.new(m.x + oy, m.y - ox), Two]
}

var td = [
[Point.new(0.1234, 0.9876), Point.new(0.8765, 0.2345), 2.0],
[Point.new(0.0000, 2.0000), Point.new(0.0000, 0.0000), 1.0],
[Point.new(0.1234, 0.9876), Point.new(0.1234, 0.9876), 2.0],
[Point.new(0.1234, 0.9876), Point.new(0.8765, 0.2345), 0.5],
[Point.new(0.1234, 0.9876), Point.new(0.1234, 0.9876), 0.0]
]
for (tc in td) {
System.print("p1: %(tc[0])")
System.print("p2: %(tc[1])")
System.print("r : %(tc[2])")
var res = circles.call(tc[0], tc[1], tc[2])
System.print("    %(res[2])")
if (res[2] == CoR0 || res[2] == Diam) {
System.print("    Center: %(res[0])")
} else if (res[2] == Two) {
System.print("    Center 1: %(res[0])")
System.print("    Center 2: %(res[1])")
}
System.print()
}
```
Output:
```p1: (0.1234, 0.9876)
p2: (0.8765, 0.2345)
r : 2
Two circles.
Center 1: (1.8631118016582, 1.9742118016582)
Center 2: (-0.86321180165819, -0.75211180165819)

p1: (0, 2)
p2: (0, 0)
r : 1
Points form a diameter and describe only a single circle.
Center: (0, 1)

p1: (0.1234, 0.9876)
p2: (0.1234, 0.9876)
r : 2
Coincident points describe an infinite number of circles.

p1: (0.1234, 0.9876)
p2: (0.8765, 0.2345)
r : 0.5
Points too far apart to form circles.

p1: (0.1234, 0.9876)
p2: (0.1234, 0.9876)
r : 0
Coincident points with r == 0 describe a degenerate circle.
Center: (0.1234, 0.9876)
```

## XPL0

An easy way to solve this: translate the coordinates so that one point is at the origin. Then rotate the coordinate frame so that the second point is on the X-axis. The circles' X coordinate is then half the distance to the second point. The circles' Y coordinates are easily seen as +/-sqrt(radius^2 - circleX^2). Now undo the rotation and translation. The method used here is a streamlining of these steps.

```include c:\cxpl\codes;

proc Circles; real Data; \Show centers of circles, given points P & Q and radius
real Px, Py, Qx, Qy, R, X, Y, X1, Y1, Bx, By, PB, CB;
[Px:= Data(0); Py:= Data(1); Qx:= Data(2); Qy:= Data(3); R:= Data(4);
if R = 0.0 then [Text(0, "Radius = zero gives no circles^M^J"); return];
X:= (Qx-Px)/2.0;  Y:= (Qy-Py)/2.0;
Bx:= Px+X;  By:= Py+Y;
PB:= sqrt(X*X + Y*Y);
if PB = 0.0 then [Text(0, "Coincident points give infinite circles^M^J"); return];
if PB > R   then [Text(0, "Points are too far apart for radius^M^J"); return];
CB:= sqrt(R*R - PB*PB);
X1:= Y*CB/PB; Y1:= X*CB/PB;
RlOut(0, Bx-X1); ChOut(0, ^,); RlOut(0, By+Y1); ChOut(0, 9\tab\);
RlOut(0, Bx+X1); ChOut(0, ^,); RlOut(0, By-Y1); CrLf(0);
];

real Tbl; int I;
[Tbl:=[[0.1234, 0.9876,    0.8765, 0.2345,    2.0],
[0.0000, 2.0000,    0.0000, 0.0000,    1.0],
[0.1234, 0.9876,    0.1234, 0.9876,    2.0],
[0.1234, 0.9876,    0.8765, 0.2345,    0.5],
[0.1234, 0.9876,    0.1234, 0.9876,    0.0]];
for I:= 0 to 4 do Circles(Tbl(I));
]```
Output:
```    1.86311,    1.97421    -0.86321,   -0.75211
0.00000,    1.00000     0.00000,    1.00000
Coincident points give infinite circles
Points are too far apart for radius
Radius = zero gives no circles
```

## Yabasic

Translation of: Liberty BASIC
```sub twoCircles (x1, y1, x2, y2, radio)
if x1 = x2 and y1 = y2 then //Si los puntos coinciden
print "Los puntos son los mismos\n"
return true
else
print "Hay cualquier numero de circulos a traves de un solo punto (", x1, ",", y1, ") de radio ", radio : print
return true
end if
end if
r2 = sqr((x1-x2)^2+(y1-y2)^2) / 2  //distancia media entre puntos
return true
end if

//si no, calcular dos centros
cx = (x1+x2) / 2 //punto medio
cy = (y1+y2) / 2
//debe moverse desde el punto medio a lo largo de la perpendicular en dd2
dd2 = sqr(radio^2 - r2^2)   //distancia perpendicular
dx1 = x2-cx           //vector al punto medio
dy1 = y2-cy
dx = 0-dy1 / r2*dd2   //perpendicular:
dy = dx1 / r2*dd2     //rotar y escalar
print " -> Circulo 1 (", cx+dy, ", ", cy+dx, ")"     //dos puntos, con (+)
print " -> Circulo 2 (", cx-dy, ", ", cy-dx, ")\n"   //y (-)
end sub

for i = 1 to 5
print "Puntos ", "(", x1, ",", y1, "), (", x2, ",", y2, ")", ", Radio ", radio
twoCircles (x1, y1, x2, y2, radio)
next
end

data 0.1234, 0.9876,    0.8765, 0.2345,    2.0
data 0.0000, 2.0000,    0.0000, 0.0000,    1.0
data 0.1234, 0.9876,    0.1234, 0.9876,    2.0
data 0.1234, 0.9876,    0.8765, 0.2345,    0.5
data 0.1234, 0.9876,    0.1234, 0.9876,    0.0```

## zkl

Translation of: C
```fcn findCircles(a,b, c,d, r){ //-->T(T(x,y,r) [,T(x,y,r)]))
delta:=(a-c).hypot(b-d);
switch(delta){	// could just catch MathError
case(0.0){"singularity"}  // should use epsilon test
case(r*2){T(T((a+c)/2,(b+d)/2,r))}
else{
if(delta > 2*r) "Point delta > diameter";
else{
md:=(r.pow(2) - (delta/2).pow(2)).sqrt();
T(T((a+c)/2 + md*(b-d)/delta,(b+d)/2 + md*(c-b)/delta,r),
T((a+c)/2 - md*(b-d)/delta,(b+d)/2 - md*(c-b)/delta,r));
}
}
}
}

data:=T(
T(0.1234, 0.9876,    0.8765, 0.2345,    2.0),
T(0.0000, 2.0000,    0.0000, 0.0000,    1.0),
T(0.1234, 0.9876,    0.1234, 0.9876,    2.0),
T(0.1234, 0.9876,    0.8765, 0.2345,    0.5),
T(0.1234, 0.9876,    0.1234, 0.9876,    0.0),
);

ppFmt:="(%2.4f,%2.4f)";
pprFmt:=ppFmt+" r=%2.1f";
foreach a,b, c,d, r in (data){
println("Points: ",ppFmt.fmt(a,b),", ",pprFmt.fmt(c,d,r));
print("   Circles: ");
cs:=findCircles(a,b,c,d,r);
if(List.isType(cs))
print(cs.pump(List,'wrap(c){pprFmt.fmt(c.xplode())}).concat(", "));
else print(cs);
println();
}```
Output:
```Points: (0.1234,0.9876), (0.8765,0.2345) r=2.0
Circles: (1.8631,1.9742) r=2.0, (-0.8632,-0.7521) r=2.0
Points: (0.0000,2.0000), (0.0000,0.0000) r=1.0
Circles: (0.0000,1.0000) r=1.0
Points: (0.1234,0.9876), (0.1234,0.9876) r=2.0
Circles: singularity
Points: (0.1234,0.9876), (0.8765,0.2345) r=0.5
Circles: Point delta > diameter
Points: (0.1234,0.9876), (0.1234,0.9876) r=0.0
Circles: singularity
```

## ZX Spectrum Basic

Translation of: Liberty BASIC
```10 FOR i=1 TO 5
30 PRINT i;") ";x1;" ";y1;" ";x2;" ";y2;" ";r
40 GO SUB 1000
50 NEXT i
60 STOP
70 DATA 0.1234,0.9876,0.8765,0.2345,2.0
80 DATA 0.0000,2.0000,0.0000,0.0000,1.0
90 DATA 0.1234,0.9876,0.1234,0.9876,2.0
100 DATA 0.1234,0.9876,0.8765,0.2345,0.5
110 DATA 0.1234,0.9876,0.1234,0.9876,0.0
1000 IF NOT (x1=x2 AND y1=y2) THEN GO TO 1090
1010 IF r=0 THEN PRINT "It will be a single point (";x1;",";y1;") of radius 0": RETURN
1020 PRINT "There are any number of circles via single point (";x1;",";y1;") of radius ";r: RETURN
1090 LET p1=(x1-x2): LET p2=(y1-y2)
1100 LET r2=SQR (p1*p1+p2*p2)/2
1110 IF r<r2 THEN PRINT "Points are too far apart (";2*r2;") - there are no circles of radius ";r: RETURN
1120 LET cx=(x1+x2)/2
1130 LET cy=(y1+y2)/2
1140 LET dd2=SQR (r^2-r2^2)
1150 LET dx1=x2-cx
1160 LET dy1=y2-cy
1170 LET dx=0-dy1/r2*dd2
1180 LET dy=dx1/r2*dd2
1190 PRINT "(";cx+dy;",";cy+dx;")"
1200 PRINT "(";cx-dy;",";cy-dx;")"
1210 RETURN```