Circles of given radius through two points

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

Exceptions

r==0.0 should be treated as never describing circles.
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.
If the points form a diameter then return two identical circles.
If the points are too far apart then no circles can be drawn.

Task detail

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

Ref

Finding the Center of a Circle from 2 Points and Radius from Math forum @ Drexel

Python

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

<lang python>from collections import namedtuple from math import sqrt

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

def circles_from_p1p2r(p1, p2, r):

   'Following explanation at http://mathforum.org/library/drmath/view/53027.html'
   if r == 0.0:
       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(0.1234, 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],))</lang>

Output:

Through points:
  Pt(x=0.1234, y=0.9876),
  Pt(x=0.8765, y=0.2345)
  and radius 2.000000
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)
  and radius 1.000000
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)
  and radius 2.000000
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)
  and radius 0.500000
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)
  and radius 0.000000
You can construct the following circles:
  ERROR: radius of zero

Tcl

Translation of: Python

<lang tcl>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]

}</lang> Demonstrating: <lang tcl>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"

}</lang>

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)