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Tangent Circles

lf the circle...

Question

lf the circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} - 6 x - 8 y + K = 0$ touch internally then $K =$

- 24

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Solution

The correct option is C $- 24$
[ From point (1) ]
$S_{1} : x^{2} + y^{2} = 4$
$C_{1} \equiv (0, 0)$ and $r_{1} = 2$
$S_{2} : x^{2} + y^{2} - 6 x - 8 y + k = 0$
$C_{2} \equiv (3, 4)$ and $r_{2} = \sqrt{25 - k}$
Given $S_{1}$ & $S_{2}$ touch internally.
then, $C_{1} C_{2} = r_{2} - r_{1}$
$\sqrt{3^{2} + 4^{2}} = \sqrt{25 - k} - 2$
$5 = \sqrt{25 - k} - 2$
$7 = \sqrt{25 - k}$
$49 = 25 - k$
$k = 25 - 49$
$k = - 24$

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