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

(3,-2)

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(2,3)

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(3,2)

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(2,3)

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Solution

The correct option is A
(3,-2)

A = (6, -6), B = (3, -7), C = (3, 3)
If O is the centre, then OA = OB = OC (radii are equal)
Let O = (x, y)

$∵$ Distance between points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ .
$\Rightarrow O A = \sqrt{(x - 6)^{2} + (y + 6)^{2}}$
$\Rightarrow O B = \sqrt{(x - 3)^{2} + (y + 7)^{2}}$
$\Rightarrow O C = \sqrt{(x - 3)^{2} + (y - 3)^{2}}$
Now, OA =OB ( $∵$ Distance from centre to any point on the circle is equal)
$\Rightarrow (x - 6)^{2} + (y + 6)^{2} = (x - 3)^{2} + (y + 7)^{2}$
$\Rightarrow x^{2} - 12 x + 36 + y^{2} + 12 y + 36 = x^{2} - 6 x + 9 + y^{2} + 14 y + 49$
$\Rightarrow x^{2} - 12 x + y^{2} + 12 y + 72 = x^{2} - 6 x + y^{2} + 14 y + 58$
$\Rightarrow x^{2} - 12 x - (x^{2} - 6 x) = y^{2} + 14 y - (y^{2} + 12 y) + 58 - 72$
$\Rightarrow - 12 x + 6 x = 14 y - 12 y - 14$
$\Rightarrow - 6 x = 2 y - 14..... (1)$

Similarly OB = OC
$\Rightarrow (x - 3)^{2} + (y + 7)^{2}$ $= (x - 3)^{2} + (y - 3)^{2}$
$\Rightarrow (y + 7)^{2}$ $= (y - 3)^{2}$
$\Rightarrow y^{2} + 14 y + 49 = y^{2} - 6 y + 9$
$\Rightarrow 20 y = - 40$
$\Rightarrow y = - 2$
Substituting the value of y in equation (1), we get;
$- 6 x = 2 y - 14$
$\Rightarrow - 6 x = - 4 - 14 = - 18$
$\Rightarrow x = 3$
Hence, x = 3, y = - 2 are the coordinates of centre.

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