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

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$x^{2} + y^{2} - 2 p x - 3 q y + (p^{2} - q^{2} - r^{2}) = 0$

$2 p x + 2 q y - (p^{2} + q^{2} + r^{2}) = 0$

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

$2 p x + 2 q y - (p^{2} - q^{2} + r^{2}) = 0$

answer is A.

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Detailed Solution

Let P(x₁,y₁) be the locus point , C(0,0) is centre of given circle.

The two circles intersect orthogonally

$\Rightarrow d^{2} = {r_{1}}^{2} + {r_{2}}^{2}$

${x_{1}}^{2} + {y_{1}}^{2} = {(x_{1} - p)}^{2} + {(y_{1} - q)}^{2} + r^{2}$

The locus of center (x₁,y₁) is

$2 p x + 2 q y - (p^{2} + q^{2} + r^{2}) = 0$

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