Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3)
Let O (x, y) be the centre of the circle. And let the points (6, - 6), (3, - 7), and (3, 3) be representing the points A, B,and C on the circumference of the circle
OA = [(x- 6)2 + (y+ 6)2]1/2
OB = [(x- 3)2 + (y+ 7)2]1/2
OC = [(x- 3)2 + (y- 3)2]1/2
However, OA = OB
[(x - 6)2 + (y + 6)2]1/2 = [(x - 3)2 + (y + 7)2]1/2
-6x – 2y + 14 = 0
3x + y = 7 (i)
Similarly, OA = OC
[(x - 6)2 + (y + 6)2]1/2 = [(x - 3)2 + (y - 3)2]1/2
-6x + 18y + 54 = 0
-3x + 9y = -27 (ii)
On adding equation (i) and (ii), we obtain
10y = - 20
y = - 2
From equation (i), we obtain
3x - 2 = 7
3x = 9
x = 3
Therefore, the centre of the circle is (3, - 2)