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

y^{2} = 16 x + 4

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x^{2} = 16 y + 4

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x^{2} = 16 y

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y^{2} = 16 x

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Solution

The correct option is D $y^{2} = 16 x$
Let S: $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ is the second circle.
$c e n t r e \equiv (- g, - f)$ .radius $= \sqrt{g^{2} + f^{2} - c}$
This circle and $x^{2} + y^{2} - 20 x + 4 = 0$ are orthogonal to each other, then by condition of orthogonality
$2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$
$\Rightarrow 2 g (- 10) + 0 = c + 4$
$- 20 g = c + 4$ ---------(1)
and circle s touch $x = 2$ line,
$\sqrt{g^{2} + f^{2} - c} = ∣ ∣ ∣ \frac{- g - 2}{1} ∣ ∣ ∣$
$g^{2} + f^{2} - c = g^{2} + 4 + 4 g$
$f^{2} - c = 4 + 4 g$ --------(2)
from (1) & (2)
$f^{2} = 4 g - 20 g$
$f^{2} = - 16 g$
$\Rightarrow (- f)^{2} = 16 (- g)$
So, locus of centre $(- g, - f)$ is
$\Rightarrow y^{2} = 16 x$

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