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

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

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

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

answer is A.

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Detailed Solution

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

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

where $a^{2} + b^{2} + 2 a g + 2 f b + c = 0$ ..(ii)

Since the circle $x^{2} + y^{2} = p^{2}$ cuts the circle (i) orthogonally.

$∴ 2 g \times 0 + 2 f \times 0 = c - p^{2} \Rightarrow c = p^{2}$ ..(iii)

Substituting the value of $c$ in (ii), we obtain

$a^{2} + b^{2} + 2 a g + 2 f b + p^{2} = 0$

Hence, the locus of $(- g, - f)$ is $a^{2} + b^{2} - 2 a x - 2 b y + p^{2} = 0$ .

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