# 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
```

## 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:       ( "radius:"       + 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)
```

## 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)		return "radius is not valid"	if !(cx1 && cy1 && cx2 && cy2)		return "no solution"	return cx1 "," cy1 " & " cx2 "," cy2}`
Examples:
`data = (0.1234 0.9876 0.8765 0.2345 2.00.0000 2.0000 0.0000 0.0000 1.00.1234 0.9876 0.1234 0.9876 2.00.1234 0.9876 0.8765 0.2345 0.50.1234 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
```

## BASIC

Works with: FreeBASIC
`Type Point    As Double x,y    Declare Property length As DoubleEnd Type Property point.length As DoubleReturn Sqr(x*x+y*y)End Property Sub circles(p1 As Point,p2 As Point,radius As Double)    Print "Points ";"("&p1.x;","&p1.y;"),("&p2.x;","&p2.y;")";", Rad ";radius    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 dist=Sqr(radius^2-lenhalf^2)/lenhalf    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;")"    PrintEnd 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=p1circles(p1,p2,2)p1=Type<Point>(.1234,.9876):p2=Type<Point>(.8765,.2345)circles(p1,p2,.5)p1=Type<Point>(.1234,.9876):p2=p1circles(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```

## 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));} void findCircles(point p1,point p2,double radius){	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);	} 	else if(separation == 2*radius)	{		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); 	} 	else if(separation > 2*radius)	{		printf("\nGiven points are farther away from each other than a diameter of a circle with radius %.4f",radius);	}    	else	{		mirrorDistance =sqrt(pow(radius,2) - pow(separation/2,2)); 		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}    };     double radii[] = {2.0,1.0,2.0,0.5,0.0};     for(i=0;i<5;i++)    {		printf("\nCase %d)",i+1);	findCircles(cases[2*i],cases[2*i+1],radii[i]);    }     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 class CirclesOfGivenRadiusThroughTwoPoints{    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(radius < 0) throw new ArgumentException("Negative radius.");        if(radius == 0) {            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);        double sqDiameter = 4 * radius * radius;        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 d = Math.Sqrt(radius * radius - sqDistance / 4);        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 = std::hypot(r, half_distance) / 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)        print(find_circles(points[i*2], points[i*2 + 1], radius[i]));}`
Output:
```Two solutions: 1.96344 2.07454 and -0.963536 -0.852436
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)))        throw new ValueException("radius of zero");     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",                     circlesFromTwoPointsAndRadius(t[])[]);        } 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:
```

## 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  endend 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             MSG\$="RADIUS IS NOT VALID"             EXIT PROCEDURE        END IF        MSG\$=STR\$(CX1)+","+STR\$(CY1)+" & "+STR\$(CX2)+","+STR\$(CY2)END PROCEDURE BEGINDATA(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   READ(PX,PY,QX,QY,RADIUS)   CIRCLE_CENTER(PX,PY,QX,QY,RADIUS->MSG\$)   PRINT(MSG\$)END FOREND 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 * blet 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 magMidC = Math.Sqrt(Math.Abs(radius * radius - halfPQ * halfPQ))    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-circuitformatting io kernel literals locals math math.distancesmath.functions prettyprint sequences strings ;IN: rosetta-code.circlesDEFER: 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) ] [email protected] ;    inline :: find-circles ( input -- circles )    input [ r>> ] [ p1>> ] [ p2>> ] tri    :> ( r p1 p2 )    p1 p2 [ [ first ] [ second ] bi ] [email protected]  :> ( x1 y1 x2 y2 )    x1 x2 y1 y2 [ + 2 / ] [email protected]             :> ( x3 y3 )    p1 p2 euclidian-distance               :> q    [ x3 y1 y2 r q ]    [ y3 x2 x1 r q ] [ circle-coords ] [email protected] :> ( 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 circlesimplicit nonedouble precision :: P1(2), P2(2), R P1 = (/0.1234d0, 0.9876d0/)P2 = (/0.8765d0,0.2345d0/)R = 2.0d0call print_centers(P1,P2,R) P1 = (/0.0d0, 2.0d0/)P2 = (/0.0d0,0.0d0/)R = 1.0d0call print_centers(P1,P2,R) P1 = (/0.1234d0, 0.9876d0/)P2 = (/0.1234d0, 0.9876d0/)R = 2.0d0call print_centers(P1,P2,R) P1 = (/0.1234d0, 0.9876d0/)P2 = (/0.8765d0, 0.2345d0/)R = 0.5d0call print_centers(P1,P2,R) P1 = (/0.1234d0, 0.9876d0/)P2 = (/0.1234d0, 0.9876d0/)R = 0.0d0call print_centers(P1,P2,R)end program circles subroutine print_centers(P1,P2,R)implicit nonedouble precision :: P1(2), P2(2), R, Center(2,2)integer :: Rescall 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)write(*,'(A10,F7.4)') 'Radius  : ', Rif(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 ifelseif(Res.lt.0) then  write(*,*) 'Incorrect value for r.'end ifwrite(*,*)end subroutine print_centers subroutine test_inputs(P1,P2,R,Res)implicit nonedouble precision :: P1(2), P2(2), R, distinteger :: Resif(R.lt.0.0d0) then  Res = -1  returnelseif(R.eq.0.0d0 .and. P1(1).eq.P2(1) .and. P1(2).eq.P2(2)) then  Res = 1  returnelseif(R.eq.0.0d0) then  Res = 2  returnelseif(P1(1).eq.P2(1) .and. P1(2).eq.P2(2)) then  Res = 3  returnelse  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 ifend ifend subroutine test_inputs subroutine find_center(P1,P2,R,Center)implicit nonedouble precision :: P1(2), P2(2), MP(2), Center(2,2), R, dm, ddMP = (P1 + P2)/2.0d0dm = 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))/dmCenter(1,2) = MP(2) + dd*(P2(1) - P1(1))/dm Center(2,1) = MP(1) + dd*(P2(2) - P1(2))/dmCenter(2,2) = MP(2) - dd*(P2(1) - P1(1))/dmend 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.        REAL R		!The specified radius.        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!
```

## 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}
```

`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 amul (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))circlePoints p q radius  | radius == 0      = Nothing  | p == q           = Nothing  | diameter < magPQ = Nothing  | otherwise        = Just (center1, center2)  where    diameter = radius * 2    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    center1  = midpoint `add` midC    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 [email protected]:, P1 BAD =: ":@:+. C 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 [email protected]:+ 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];            double r = radii[i];            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)}` :        'Radius Zero';      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: 0,1Test: 2: p1: 0.1234,0.9876, p2: 0.1234,0.9876, r:2 Result: Coincident point. Infinite solutionsTest: 3: p1: 0.1234,0.9876, p2: 0.8765,0.2345, r:0.5 Result: No intersection. Points further apart than circle diameterTest: 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::Tend immutable Circle{T<:FloatingPoint}    c::Point{T}    r::TendCircle{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))    endend `
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]                        val r  = radii[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)
```

## Liberty BASIC

` '[RC] Circles of given radius through two pointsfor i = 1 to 5    read x1, y1, x2, y2,r    print i;") ";x1, y1, x2, y2,r    call twoCircles x1, y1, x2, y2,rnextend 'p1                p2           rdata 0.1234, 0.9876,    0.8765, 0.2345,    2.0data 0.0000, 2.0000,    0.0000, 0.0000,    1.0data 0.1234, 0.9876,    0.1234, 0.9876,    2.0data 0.1234, 0.9876,    0.8765, 0.2345,    0.5data 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        2There are any number of circles via single point (0.1234,0.9876) of radius 24) 0.1234     0.9876        0.8765        0.2345        0.5Points are too far apart (1.06504423) - there are no circles of radius 0.55) 0.1234     0.9876        0.1234        0.9876        0It 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 function findCircles(p1, p2, radius)    local seperation = distance(p1, p2)    if seperation == 0.0 then        if radius == 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    elseif seperation == 2*radius then        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)    elseif seperation > 2*radius then        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.")        print("Circle C1 with center ("..c1.x..", "..c1.y.."), radius "..radius)        print("Circle C2 with center ("..c2.x..", "..c2.y.."), radius "..radius)    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 }for i=1, #radii do    print("Case "..i)    findCircles(cases[i*2-1], cases[i*2], radii[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

`Off[Solve::ratnz];circs::invrad = "The radius is invalid.";circs::equpts = "The given points (`1`, `2`) are equivalent.";circs::dist =   "The given points (`1`, `2`) and (`3`, `4`) are too far apart for \radius `5`.";circs[_, _, 0.] := Message[circs::invrad];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:
```In[2]:= circs[{.1234, .9876}, {.8765, .2345}, 2.]

Out[2]= {{-0.863212, -0.752112}, {1.86311, 1.97421}}

In[3]:= circs[{.1234, .9876}, {.1234, .9876}, 2.]

circs::equpts: The given points (0.1234`, 0.9876`) are equivalent.

In[4]:= circs[{.1234, .9876}, {.8765, .2345}, .5]

circs::dist: The given points (0.1234`, 0.9876`) and (0.8765`, 0.2345`) are too

In[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],  load(to_poly_solve),  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	П4ИП2	ИП0	-	x^2	ИП3	ИП1	-	x^2	+	КвКор	П5ИП0	ИП2	+	2	/	П6	ИП1	ИП3	+	2	/	П7ИП4	x^2	ИП5	2	/	x^2	-	КвКор	ИП5	/	П8ИП6	ИП1	ИП3	-	ИП8	*	П9	+	ПAИП6	ИП9	-	ПCИП7	ИП2	ИП0	-	ИП8	*	П9	+	ПBИП7	ИП9	-	ПDИП5	x#0	97	8	4	ИНВ	С/ПИП4	2	*	ИП5	-	ПE	x#0	97	ИПB	ИПA	8	5	ИНВ	С/ПИПE	x>=0	97	8	3	ИНВ	С/ПИПD	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;FROM Terminal IMPORT WriteString,WriteLn,ReadChar; 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);    WriteString(" with radius ");    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 TRUEEXCEPT    GetMessage(buf);    WriteString(buf);    WriteLn;    WriteLn;    RETURN FALSEEND 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);     ReadCharEND 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,    "radius of zero")  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))  # One answer  result.c1 = (p3.x - d*dy/q, p3.y + d*dx/q, abs(r))  # The other answer  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:

## 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.```

## 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

## Perl 6

`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.perl, ': ', 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```

## Phix

`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.
```

## 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 namedtuplefrom 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:        raise ValueError('radius of zero')    (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)    # One answer    c1 = Cir(x = x3 - d*dy/q,             y = y3 + d*dx/q,             r = abs(r))    # The other answer    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)) `

## REXX

Translation of: XPL0

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

`/*REXX program finds  two circles  with a  specific radius  given two (X,Y)  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    0say '     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 f(p1)  f(p2)  f(p3)  f(p4)       center(r/1,9)      "───► "        2circ(@.j)       end      /*j*/exit                                             /*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  f(bx-x1)   f(by+y1)   f(bx+x1)   f(by-y1)/*──────────────────────────────────────────────────────────────────────────────────────*/f:     f=right( format( arg(1), , 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:
```     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 = 0.1234y1 = 0.9876x2 = 0.8765y2 = 0.2345r = 2.0see "1 : " + x1 + " " + y1 + " " + x2 + " " + y2 + " " + r + nltwocircles(x1, y1, x2, y2, r) x1 = 0.0000y1 = 2.0000x2 = 0.0000y2 = 0.0000r = 1.0see "2 : " + x1 + " " + y1 + " " + x2 + " " + y2 + " " + r + nltwocircles(x1, y1, x2, y2, r) x1 = 0.1234y1 = 0.9876x2 = 0.1234y2 = 0.9876r = 2.0see "3 : " + x1 + " " + y1 + " " + x2 + " " + y2 + " " + r + nltwocircles(x1, y1, x2, y2, r) x1 = 0.1234y1 = 0.9876x2 = 0.8765y2 = 0.2345r = 0.5see "4 : " + x1 + " " + y1 + " " + x2 + " " + y2 + " " + r + nltwocircles(x1, y1, x2, y2, r) x1 = 0.1234y1 = 0.9876x2 = 0.1234y2 = 0.9876r= 0.0see "5 : " + x1 + " " + y1 + " " + x2 + " " + y2 + " " + r + nltwocircles(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)].uniqend # 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  putsend`
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    read x1, y1, x2, y2,rhtml "<tr align=right><td>";i;"</td><td>";x1;"</td><td>";y1;"</td><td>";x2;"</td><td>";y2;"</td><td>";r;"</td>"    gosub [twoCircles]nexthtml "</table>"end 'p1                p2           rdata 0.1234, 0.9876,    0.8765, 0.2345,    2.0data 0.0000, 2.0000,    0.0000, 0.0000,    1.0data 0.1234, 0.9876,    0.1234, 0.9876,    2.0data 0.1234, 0.9876,    0.8765, 0.2345,    0.5data 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.FunSuiteimport 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 {      Right(circlesThruPoints(p1, p2, radius))    } 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 y-coord cadr)  (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))             (equal-points? (car pts) (cadr 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      if radius = 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 " <&              radius digits 4);     elsif separation > 2.0 * radius then      writeln("Given points are farther away from each other than a diameter of a circle with radius " <&              radius digits 4);    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 " <&              radius digits 4);      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 " <&              radius digits 4);    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: Perl 6
`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  |    +-----------------+`

## 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}]    # The two answers    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 iEnd 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        If System.Diagnostics.Debugger.IsAttached Then Console.ReadKey()    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 ArgumentException("Negative radius.")        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,            sqDiam As Double = 4 * rad * rad        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 StructureEnd 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 DoubleCLOSE DATABASES ALLSET FIXED ONSET DECIMALS TO 4CLEARCREATE 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)    rr = rad    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 Stringctr = 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.nLengthPrintPoints(op1, op2, r)IF r < lenhalf    ? "Cannot solve for these parameters."    RETURNENDIFIF lenhalf = 0    ? "Points are coincident."    RETURNENDIFdist = SQRT(r^2 - lenhalf^2)/lenhalfrot = 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? crot.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? cENDPROC PROCEDURE PrintPoints(op1 As point, op2 As point, r As Double)LOCAL lcTxt As StringTEXT TO lcTxt TEXTMERGE NOSHOW PRETEXT 3    Points (<<op1.xc>>,<<op1.yc>>), (<<op2.xc>>,<<op2.yc>>) Radius <<r>>.ENDTEXT? lcTxtENDPROC	 DEFINE CLASS point As Customxc = 0yc = 0nLength = 0 PROCEDURE InitDODEFAULT()ENDPROC PROCEDURE SetPoints(tnx As Double, tny As Double)THIS.xc = tnxTHIS.yc = tnyTHIS.nLength = THIS.GetLength()ENDPROC FUNCTION GetLength()RETURN SQRT(THIS.xc*THIS.xc + THIS.yc*THIS.yc)ENDFUNC ENDDEFINE `
Output:
```Points (0.1234,0.9876), (0.8765,0.2345) Radius 2.0000.
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.
```

## 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 radiusreal 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
```

## 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 520 READ x1,y1,x2,y2,r30 PRINT i;") ";x1;" ";y1;" ";x2;" ";y2;" ";r40 GO SUB 100050 NEXT i60 STOP 70 DATA 0.1234,0.9876,0.8765,0.2345,2.080 DATA 0.0000,2.0000,0.0000,0.0000,1.090 DATA 0.1234,0.9876,0.1234,0.9876,2.0100 DATA 0.1234,0.9876,0.8765,0.2345,0.5110 DATA 0.1234,0.9876,0.1234,0.9876,0.01000 IF NOT (x1=x2 AND y1=y2) THEN GO TO 10901010 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)/21110 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)/21130 LET cy=(y1+y2)/21140 LET dd2=SQR (r^2-r2^2)1150 LET dx1=x2-cx1160 LET dy1=y2-cy1170 LET dx=0-dy1/r2*dd21180 LET dy=dx1/r2*dd21190 PRINT "(";cx+dy;",";cy+dx;")"1200 PRINT "(";cx-dy;",";cy-dx;")"1210 RETURN`