If a circle 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 b y - (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 equation of circle be
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
it cuts the circle $x^{2} + y^{2} = 4$ orthogonally if
$2 g \times 0 + 2 f \times 0 = c - 4 \Rightarrow c = 4$
$∴$ The equation of the circle is
$x^{2} + y^{2} + 2 g x + 2 f y + 4 = 0$
Since, it passes through the point (a, b),
$a^{2} + b^{2} - 2 x a - 2 y b + 4 = 0 \Rightarrow 2 a x + 2 b y - (a^{2} + b^{2} + 4) = 0$

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x^{2} + y^{2} = 4

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