The locus of the centres of the circles which cut the circles $x^{2} + y^{2} + 4 x - 6 y + 9 = 0$ and $x^{2} + y^{2} - 5 x + 4 y - 2 = 0$ orthogonally is

9 x + 10 y - 7 = 0

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x - y + 2 = 0

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9 x - 10 y + 11 = 0

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9 x + 10 y + 7 = 0

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Solution

The correct option is C $9 x - 10 y + 11 = 0$
Let the centre of circle be $(- g, - f)$
Using condition of orthogonality
$2 (g_{1} g_{2} + f_{1} f_{2}) = C_{1} + C_{2}$
$2 (2 g - 3 f) = 9 + C . . . . . . . . (i)$
$2 (- \frac{5 g}{2} + 2 f) = - 2 + C . . . . . . . . . . (i i)$
Subtract $(i i)$ from $(i)$
$2 [\frac{9 g}{2} - 5 f] = 11 \Rightarrow 9 g - 10 f = 11$
Now for locus of centre replace $(- g)$ by $x$ and $(- f)$ by $y$
$\Rightarrow$ $9 x - 10 y + 11 = 0$

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