If a circles passes through the point $(a, b)$ and cuts the circle $x^{2} + y^{2} = 4$ orthogonally, then the locus of its centre is

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

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

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2 a x + 2 b y + (a^{2} + b^{2} + 4) = 0

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2 a x - 2 a b - (a^{2} + b^{2} + 4) = 0

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Solution

The correct option is B $2 a x + 2 b y - (a^{2} + b^{2} + 4) = 0$
Let the variables of the two circles is
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0 \to (i)$

If passes through $(x, y)$
if passes through $(a, b)$

Therefore,
$\begin{matrix} a^{2} + b^{2} + 2 g a + 2 f b + c = 0 . . . . . . . . (i i) L e t x^{2} + y^{2} = 4 o r t h o g o n a l l y 2 (g \times 0 + f \times 0) = c - 4 c = 4 f r o m e q u a t i o n \to (i i) a^{2} + b^{2} + 2 g a + 2 f b + 4 = 0 l o c u s o f c e n t r e (- g, - f) i s a q^{2} + b^{2} - 2 a x - 2 b y + 4 = 0 2 a x + 2 b y = a^{2} + b^{2} + 4 2 a x + 2 b y - (a^{2} + b^{2} + 4) = 0 \end{matrix}$

Hence, this is the answer.

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