Find the centre of a circle passing through the points (6, -6), (3, -7) and (3, 3).

Let O(x,y) be the center of the circle and A(6, -6), B(3, -7) and C(3, 3) be the points on the circumference of the circle.
∴OA = √((x₁-x₂)² + (y₁-y₂)²)
⇒OA = √(x-6)² + (y+6)²
⇒OB = √(x-3)² + (y+7)²
⇒OC = √(x-3)² + (y-3)²
Since radii of the circle are equal
∴OA = OB ⇒ √(x-6)² + (y+6)² = √(x-3)² + (y+7)²
⇒x² + 36 - 12x + y² + 36 + 12y = x² + 9 - 6x + y² + 49 + 14y
⇒-6x - 2y + 14 = 0
⇒3x + y = 7 (i)
Similarly, OA = OC ⇒ √(x-6)² + (y+6)² = √(x-3)² + (y-3)²
⇒x² + 36 - 12x + y² + 36 + 12y = x² + 9 - 6x + y² + 9 - 6y
⇒-6x + 18y + 54 = 0
⇒-3x + 9y = -27 (ii)
Adding (i) and (ii)
⇒10y = -20
⇒y = -2
Substitute the value of y in (i)
⇒3x + (-2) = 7
⇒3x = 9
⇒x = 3
∴The center of the circle is (3, -2).