Let $S_{1} : x^{2} - 10 x + y^{2} + 16 = 0$ and $S_{2}$ be image of $S_{1}$ w.r.t line $x - y = 0$ and $S_{3}$ be image of $S_{2}$ w. r. t. $y = 0.$ If S’ be the circle which cuts all three circles orthogonally, then :

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Equation of $S_{3}$ is $x^{2} + y^{2} + 10 y + 16 = 0$

Equation of circle S’ is $x^{2} + y^{2} = 16$

Equation of $S_{2}$ is $x^{2} + y^{2} - 10 y + 16 = 0$

Radius of director circle of S’ is $8 \sqrt{2}$

answer is A, B, C.

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Detailed Solution

$∴$ Centre of $S_{1} = (5, 0)$ and radius $r_{1} = 3$
$∴$ Centre of $S_{2} = (0, 5)$ and radius $r_{2} = 3$
And Centre of $S_{3} = (0, - 5)$ and radius $r_{2} = 3$
$∴$ Radical centre of $S_{1}, S_{2}$ and $S_{3}$ will be $(0, 0)$ Length of tangent from (0, 0) upon $S_{1}$ or $S_{2}$ or $S_{3} = 4$
$∴$ Equation of S’ will be $\Rightarrow x^{2} + y^{2} = 16$ and radius =4