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

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Coordinates of points on a circle are A(6, -6), B(3, -7) and C(3, 3). 

Let the coordinates of the centre of the circle be O(x, y) 

Using distance formula = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Since the distance of the points A, B and C will be equal from the center, therefore 

⇒ OA = OC

\(\sqrt{(x - 6)^2 + (y + 6)^2}\) = \(\sqrt{(x - 3)^2 + (y - 3)^2}\)

On squaring both sides, we get

(x - 6)2 + (y + 6)2 = (x - 3)2 + (y - 3)2

x2 - 12x + 36 + y2 + 12y + 36 

= x2 - 6x + 9 + y2 - 6y + 9

x - 3y = 9........(1)

Similarly, OA = OB

\(\sqrt{(x - 6)^2 + (y + 6)^2}\) = \(\sqrt{(x - 3)^2 + (y + 7)^2}\)

On squaring both sides, we get

(x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2

x2 - 12x + 36 + y2 + 12y + 36 

= x2 - 6x + 9 + y2 + 14y + 49

3x + y + 7.......(1)

Solving eqn (1) and (2), we get 

x = 3; y = - 2 

Coordinates of circum center are (3, - 2)

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