The equation of the circle which passing through origin and cuts the circles $x^{2} + y^{2} - 4 x - 6 y - 3 = 0$ and $x^{2} + y^{2} - 8 y + 12 = 0$ orthogonally is

x^{2} + y^{2} + 6 x - 3 y = 0

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x^{2} + y^{2} + 7 x + y = 0

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x^{2} + y^{2} + x + 7 y = 0

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None of these

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Solution

The correct option is A $x^{2} + y^{2} + 6 x - 3 y = 0$
Let the required circle be $x^{2} + y^{2} + 2 g x + 2 f c + c = 0$ ...(1)
As the circle passes through $(0, 0)$ , we get $c = 0$ .
This circle is orthogonal to two given circles, we apply the condition $2 g g_{1} + 2 f f_{1} = c + c_{1}$ to get
$2 g (0) - 8 f + c + 12$ and $- 4 g - 6 f = c - 3$
$\Rightarrow - 8 f = 12$ and $- 4 g - 6 f = 3$
$\Rightarrow f = \frac{- 3}{2}$ , $g = 3$
From (1), we get $x^{2} + y^{2} + 6 x - 3 y = 0$

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