Circles of given radius through two points

Circles of given radius through two points is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

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.
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.
4. 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