Finding the Center of a Circle from 2 Points

Date: 06/01/99 at 08:24:55
From: John
Subject: Need to find centre of circle from X, Y, and r

If you have 2 points on a circle, both X and Y figures are known, and 
a radius is given, how is possible to find the centre of the circle?

Date: 06/01/99 at 10:07:40
From: Doctor Rob
Subject: Re: Need to find centre of circle from X, Y, and r

Using the two given points as centres, draw two circles, each with the 
given radius. The places where those two circles intersect are the 
possible centres of the original circle. Algebraically, if the two 
points are (a,b) and (A,B), and the radius is r, you can solve the two 
equations

   (x-a)^2 + (y-b)^2 = r^2,
   (x-A)^2 + (y-B)^2 = r^2,

simultaneously. (Hint: Subtract one from the other, and you'll have a 
linear equation as a result. Solve that for one of the variables, and 
substitute in either of the above two equations. That will give you a 
quadratic equation in the other variable.)

This works because the distance from the centre of the circle to each 
of (a,b) and (A,B) is r.

Example:  Find the center of the circle passing through (1,4) and 
(5,1) with radius 13/2.

   (x-1)^2 + (y-4)^2 = (13/2)^2,
   (x-5)^2 + (y-1)^2 = (13/2)^2,  expanding these we get

      x^2 -  2*x +  1 + y^2 - 8*y + 16 = (13/2)^2,
 (-)  x^2 - 10*x + 25 + y^2 - 2*y +  1 = (13/2)^2,
      --------------------------------------------  subtracting these
             8*x - 24       - 6*y + 15 = 0,
   
                                     y = (8*x-9)/6,  substituting this


                         (x-5)^2 + ([8*x-15]/6)^2 = (13/2)^2,
   x^2 - 10*x + 25 + (16/9)*x^2 - (20/3)*x + 25/4 = 169/4,
                       (25/9)*x^2 - (50/3)*x - 11 = 0,
                              25*x^2 - 150*x - 99 = 0,
                             (5*x - 33)*(5*x + 3) = 0,  so

   x = 33/5   or  x = -3/5,
   y = 73/10  or  y = -23/10.

Thus the center of the circle could be at either (33/5,73/10) or 
(-3/5,-23/10). You can check that these points are 13/2 units away 
from both (1,4) and (5,1).

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/