If the circle $x^{2} + y^{2} + 8 x - 4 y + c = 0$ touches the circle $x^{2} + y^{2} + 2 x + 4 y - 11 = 0$ externally and cuts the circle $x^{2} + y^{2} - 6 x + 8 y + k = 0$ orthogonally then $k =$

59

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- 59

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19

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Solution

The correct option is B $- 59$
Let $s_{1} : x^{2} + y^{2} + 8 x - 4 y + c = 0$
$∴$ centre is $4 : (- 4, 2)$ and radius $= \sqrt{20 - c}$
and $s_{2} : x^{2} + y^{2} + 2 x + 4 y - 11 = 0$
centre is $C_{2} : (- 1, - 2)$ and $r_{2} = \sqrt{5 + 11} = 4$
$∵$ these two circle one touching each other
$c_{1} c_{2} = r_{1} + r_{2}$
$\sqrt{9 + 16} = \sqrt{20 - c} + 4$
$\Rightarrow 5 - 4 = \sqrt{20 - c}$
$\Rightarrow c = 19$
$s_{3} : x^{2} + y^{2} - 6 x + 8 y + k = 0$
$∴$ centre $c_{3} : (3, - 4) r_{2} = \sqrt{25 - k}$
Since, $s_{1}$ and $s_{3}$ cuts orthogonally,
$r_{1}^{2} + r_{3}^{2} = (c_{1} c_{3})^{2}$
Hence, $(20 - c) + (25 - k) = (\sqrt{45 + 36})^{2}$
$\Rightarrow k = 26 - 85 = - 59$

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