Let A = (6, -6), B = (3, -7), C = (3, 3) are the points on a circle.

If O is the centre, then OA = OB = OC (radii are equal)

If O = (x, y) then

OA = \(\sqrt{(x-6)^2 + (y+6)^2}\)

OB = \(\sqrt{(x-3)^2 + (y+7)^2}\)

OC = \(\sqrt{(x-3)^2 + (y-3)^2}\)

OA = OB, we have

After simplifying above, we get -6x = 2y – 14 ….(1)

OB = OC

(x – 3)² + (y + 7)² = (x – 3)² + (y – 3)²

(y + 7)² = (y – 3)²

y² + 14y + 49 = y² – 6y + 9

20y =-40

or y = -2

Substituting the value of y in equation (1), we get;

-6x = 2y – 14

-6x = -4 – 14 = -18

x = 3

Hence, the centre of the circle located at point (3,-2).