A circle is in the form
(x - h)^2 + (y - k)^2 = r^2
where (h, k) is the center and r is the radius
Substituting known points in the circle, we have
[1] (2 - h)^2 + (8 - k)^2 = r^2
[2] (-4 - h)^2 + (6 - k)^2 = r^2
[3] (1 - h)^2 + (-1 - k)^2 = r^2
Since they are all equal to r^2, we can interchange the equations
[1] = [2]
(2 - h)^2 + (8 - k)^2 = (-4 - h)^2 + (6 - k)^2
=> (4 - 4h + h^2) + (64 -16k + k^2) = (16 - 8h + h^2) + (36 - 12k + k^2)
[1] = [3]
(2 - h)^2 + (8 - k)^2 = (1 - h)^2 + (-1 - k)^2
=> (4 - 4h + h^2) + (64 -16k + k^2) = (1 -2h + h^2) + (1 + 2k + k^2)
Move all to the left side and simplify
[1] = [2]
=> (4 - 4h) + (64 - 16k) - (16 - 8h) - (36 - 12k)= 0
=> 16 + 8h - 4k = 0
=> 8 + 4h - 2k = 0
[1] = [3]
=> (4 - 4h) + (64 - 16k) - (1 - 2h) - (1 + 2k) = 0
=> 66 - 2h - 18k = 0
[1'] 8 + 4h - 2k = 0
[2'] 66 - 2h - 18k = 0
Eliminate one of the variables by scaling any of the equations
such that the variable will have the same coefficient (different sign is OK) as with the other equation
[1'] 8 + 4h - 2k = 0
[2'] 132 - 4h - 18k = 0
Add (or subtract if different sign)
140 - 20k = 0
140 = 20k
k = 7
[1'] 8 + 4h -2(7) = 0
4h = 6
h = 3/2
the circle is centered at (3/2, 7)
