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)#