The locus of the centre of the circle which cuts the circle $x^{2} + y^{2} - 20 x + 4 = 0$ orthogonally and touches the line $x = 2$ is

x^{2} = 16 y

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y^{2} = 4 x

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y^{2} = 16 x

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x^{2} = 4 y

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Solution

The correct option is C $y^{2} = 16 x$
If two circles in their standard equation cut each other orthogonally, we get the condition $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$
We have $g_{1} = - 10, f_{1} = 0, c_{1} = 4$
$\Rightarrow - 20 g_{2} + 0 = 4 + c_{2}$ ...(1)
Also, since the circle touches the line $x = 2,$ distance of the center from the line equals radius.
$\Rightarrow | - g_{2} - 2 | = \sqrt{g_{2}^{2} + f_{2}^{2} - c_{2}}$
$\Rightarrow g_{2}^{2} + 4 g_{2} + 4 = g_{2}^{2} + f_{2}^{2} - c_{2}$
$\Rightarrow 4 g_{2} + 4 = f_{2}^{2} - c_{2}$ ...(2)
Adding equations (1) and (2), we get
$- 16 g_{2} = f_{2}^{2}$
Since $x = - g_{2}, y = - f_{2},$ the locus becomes $y^{2} = 16 x$

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