Center of a Circle from Circumference Points

Date: 06/25/99 at 16:03:12
From: Jared Koch
Subject: Center point of a circle

I have a circle/arc defined by 2 or more point triples (x1,y1,z1), 
(x2,y2,z2), ... on its circumference. I also know its radius. How do 
I figure its center point (Xc,Yc,Zc)?

Date: 06/26/99 at 16:11:57
From: Doctor Anthony
Subject: Re: Center point of a circle

You will need three points on the circumference to fix the plane of 
the circle.

To avoid too much symbolic algebra I will take three arbitrary points 
as being points on the circumference, and you will be able to adapt 
the method to particular cases or to a general formula if you wish.

Suppose the three points are (1, -2, 3), (3, 2, 4), (3, 1, 5).

First find the equation of the plane containing these three points

Let the equation of the plane be A(x-1) + B(y+2) + C(z-3) = 0

So we have        A(x-1) + B(y+2) + C(z-3) = 0
point (3,2,4)     A(2)   + B(4)   + C(1)   = 0
point (3,1,5)     A(2)   + B(3)   + C(2)   = 0

The equation of the plane is given by the equation

     |x-1   y+2   z-3|
     | 2     4     1 | = 0
     | 2     3     2 |

     (x-1)(5) - (y+2)(2) + (z-3)(-2) = 0     

            5x - 5 - 2y - 4 - 2z + 6 = 0

                        5x - 2y - 2z = 3    

We know that the centre of the circle will lie on this plane.

Other equations will be obtained by equating the distance of the 
centre, whose coordinates will be (a, b, c) from the three given 
points on the circumference.

    (1, -2, 3), (3, 2, 4), (3, 1, 5)

The distance^2 to each of these points is

       (a-1)^2 + (b+2)^2 + (c-3)^2
     = (a-3)^2 + (b-2)^2 + (c-4)^2
     = (a-3)^2 + (b-1)^2 + (c-5)^2

       a^2 -2a + b^2 + 4b + c^2 -6c + 14 
     = a^2 -6a + b^2 - 4b + c^2 -8c + 29
     = a^2 -6a + b^2 - 2b + c^2 -10c + 35

Canceling a^2, b^2 and c^2 across these three expressions we get

       -2a + 4b - 6c + 14
     = -6a - 4b - 8c + 29
     = -6a - 2b -10c + 35

Taking the first two of these we get

      4a + 8b + 2c = 15     and taking the second two
     0.a - 2b + 2c =  6     and putting in equation of plane
      5a - 2b - 2c =  3     

So we have three equations in three unknowns a, b, c to find the 
centre of the circle.

The solutions are a = 21/11, b = 9/66, c = 69/22, and so the centre of 
the circle has coordinates (21/11, 9/66, 69/22).

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