Let the equation of the circle be
(x-h)^2 + (y-k)^2 = r^2,
and substitute the three known points, getting 3 equations in 3
unknowns h, k, and r:
(x1-h)^2 + (y1-k)^2 = r^2
(x2-h)^2 + (y2-k)^2 = r^2
(x3-h)^2 + (y3-k)^2 = r^2
which you can solve simultaneously. First subtract the third equation
from the other two, thus eliminating r^2, h^2, and k^2. That will
leave you with 2 simultaneous linear equations in h and k to solve.
This you can do as long as the 3 points are not collinear. Then those
values of h and k can be used in the first equation to find the
radius:
r = sqrt[(x1-h)^2 + (y1-k)^2].
Example: Suppose a circle passes through the points (4,1), (-3,7), and
(5,-2). Then we know that:
(h-4)^2 + (k-1)^2 = r^2
(h+3)^2 + (k-7)^2 = r^2
(h-5)^2 + (k+2)^2 = r^2
Subtracting the first from the other two, you get:
(h+3)^2 - (h-4)^2 + (k-7)^2 - (k-1)^2 = 0,
(h-5)^2 - (h-4)^2 + (k+2)^2 - (k-1)^2 = 0,
h^2 + 6h + 9 - h^2 + 8h - 16 + k^2 - 14k + 49 - k^2 + 2k - 1 = 0
h^2 - 10h + 25 - h^2 + 8h - 16 + k^2 + 4k + 4 - k^2 + 2k - 1 = 0
14h - 12k + 41 = 0
-2h + 6k + 12 = 0
10h + 65 = 0
30h + 125 = 0
h = -13/2
k = -25/6
Then
r = sqrt[(4+13/2)^2 + (1+25/6)^2]
= sqrt[4930]/6
Therefore the equation of the circle is:
(x+13/2)^2 + (y+25/6)^2 = 4930/36
I learnt this very same problem from a site called http://www.mathforum.org/
Hope this helps,
Best of luck