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 $|k + 3 c| =$

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Detailed Solution

Use $C_{1} C_{2} = r_{1} + r_{2}$ and $d^{2} = r_{1}^{2} + r_{2}^{2}$

$\begin{array}{rcl} C_{1} C_{2} & = & D i s \tan c e b e t w e e n (- 4, 2) a n d (- 1, - 2) \\ = & \sqrt{9 + 16} \\ = & 5 \end{array}$

$\begin{array}{rcl} \sqrt{20 - c} + \sqrt{1 + 4 + 11} & = & 5 \\ \sqrt{20 - c} & = & 1 \\ 20 - c & = & 1 \\ c & = & 19 \end{array}$

Guven that $x^{2} + y^{2} + 8 x - 4 y + c = 0$ intersect the other circle $x^{2} + y^{2} - 6 x + 8 y + k = 0$ orthogonally

hence,

$\begin{array}{rcl} d^{2} & = & {r_{1}}^{2} + {r_{2}}^{2} \\ {(D i s \tan c e b e t w e e n (- 4, 2) a n d (3, - 4))}^{2} & = & {(\sqrt{16 + 4 - 19})}^{2} + {(\sqrt{9 + 16 - k})}^{2} \\ 49 + 36 & = & 1 + 25 - k \\ 59 & = & - k \\ k & = & - 59 \end{array}$