Q. 34.2( 140 Votes )

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

Answer :

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(x_{1}, y_{1}) and B(x_{2}, y_{2}) is

AB = √((x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{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

10*y* = - 20

*y* = - 2

From equation (i), we obtain

3*x* - 2 = 7

3*x* = 9

*x* = 3

Therefore, the centre of the circle is (3, - 2)

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