Locus of the centre of all circles passing through $(2, 4)$ and cutting the circle $x^{2} + y^{2} = 1$ orthogonally is

4 x - 8 y = 11

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4 x + 8 y = 21

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8 x + 4 y = 11

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4 x - 8 y = 21

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Solution

The correct option is B $4 x + 8 y = 21$
Let counter of circle is $c (h, k)$
$\Rightarrow$ Equation of circle is
$x^{2} + y^{2} - 2 h x - 2 k y + c = 0$
$c$ is a constant
This circles is orthogonal to $x^{2} + + y^{2} - 1 = 0$
$\Rightarrow 0 + 0 = c - 1$
$\Rightarrow c = 1$
( $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$ condition of orthogonal)
$\Rightarrow$ circle is $x^{2} + y^{2} - 2 h x - z k y + 1 = 0$
This circle $p$ as through $(2, 4)$
$\Rightarrow 4 + 16 + 1 - 4 h - 8 k = 0$
$\Rightarrow 4 x + 8 y = 21$

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