If a circle passes through the point $(a, b)$ and cuts the circle $x^{2} + y^{2} = p^{2}$ orthogonally, then the equation of the locus of its centre is:

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$2 a x + 2 b y - (a^{2} + b^{2} + p^{2}) = 0$

$2 a x + 2 b y - (a^{2} - b^{2} + p^{2}) = 0$

$x^{2} + y^{2} - 2 a x - 3 b y + (a^{2} - b^{2} - p^{2}) = 0$

$x^{2} + y^{2} - 3 a x - 4 b y + (a^{2} + b^{2} - p^{2}) = 0$

answer is A.

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Detailed Solution

Let the equation of the circle through $(a, b)$ be

$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ …….(1)

$∴$ $a^{2} + b^{2} + 2 g a + 2 f b + c = 0$ ………(2)

If circle (1) cuts the circle $x^{2} + y^{2} - p^{2} = 0$ , orthogonally, then

$2 g .0 + 2 f .0 = c - p^{2} \Rightarrow c = p^{2}$ .

If $C (h, k)$ is the centre of the circle (1), then

$h = - g, k = - f \Rightarrow g = - h, f = - k$ .

Substituting in (2), we get

$a^{2} + b^{2} - 2 a h - 2 b k + p^{2} = 0$

or $2 a h + 2 b k - (a^{2} + b^{2} + p^{2}) = 0$

$∴$ Locus of $C (h, k)$ is

$2 a x + 2 b y - (a^{2} + b^{2} + p^{2}) = 0$ .

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