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

The figure for the question is: 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

By distance formula, Distance between two points A(x1, y1) and B(x2, y2) is

AB = √((x2 - x1)2 + (y2 - y1)2)

Then,

OA = √[(x- 6)2 + (y+ 6)2]

OB = √[(x- 3)2 + (y+ 7)2]

OC = √[(x- 3)2 + (y- 3)2]

As OA, OB and OC are radii of same circle, 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)

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