In the Fig., two circles touch each other internally at a point A. The radius of the smaller circle with centre M is 5cm. The smaller circle passes through the centre N of the larger circle. The tangent to the smaller circle drawn through C intersects the larger circle in point D. Then CD is

\frac{20 \sqrt{2}}{3}

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\frac{80 \sqrt{2}}{3}

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\frac{10 \sqrt{2}}{3}

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\frac{40 \sqrt{2}}{3}

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Solution

The correct option is D $\frac{40 \sqrt{2}}{3}$
$G i v e n - A c i r c l e C_{1} o f r a d i u s 5 c m t o u c h e s a n o t h e r b i g g e r c i r c l e C_{2} i n t e r n a l l y a t A . C_{1} p a s s e s t h r o u g h t h e c e n t r e M o f C_{2} . T h e t a n g e n t C D t o u c h e s C_{1} a t P . C D m e e t s C_{2} a t D . T o f i n d o u t - C D = ? S o l u t i o n - A D i s j o i n e d . C A i s t h e d i a m e t e r o f C_{2} s i n c e N i s t h e c e n t r e o f C_{2} . N o w A N = 2 M N = 2 \times 5 c m = 10 c m B u t C N = A N (r a d i i o f t h e s a m e c i r c l e) . S o C N = 10 c m a n d C M = C N + M N = (10 + 5) c m = 15 c m . A l s o C A = 2 \times A N = 2 \times 10 c m = 20 c m . A g a i n ∠ C P M = 90^{o} s i n c e t h e r a d i u s, m e e t i n g t h e p o i n t o f c o n t a c t o f t h e t a n g e n t t o a c i r c l e, m a k e s 90^{o} a n g l e w i t h t h e t a n g e n t . ∴ Δ C P M i s a r i g h t o n e w i t h C M a s h y p o t e n u s e . ∴ C P = \sqrt{{C M}^{2} - {P M}^{2}} = \sqrt{15^{2} - 5^{2}} c m = 10 \sqrt{2} c m . A l s o ∠ A D C = 90^{o} s i n c e i t i s a n a n g l e i n t h e s e m i c i r c l e . ∴ Δ A D C i s a r i g h t o n e w i t h C A a s h y p o t e n u s e (a p p l y i n g P y t h a g o r a s t h e o r m) . N o w b e t w e e n Δ A D C & Δ C P M w e h a v e ∠ A D C = 90^{o} = ∠ C P M, ∠ M C P c o m m o n . ∴ Δ A D C & Δ C P M a r e s i m i l a r . i . e \frac{C D}{C A} = \frac{C P}{C M} ⟹ C D = \frac{C P}{C M} \times C A = \frac{10 \sqrt{2}}{15} \times 20 c m ⟹ C D = \frac{40 \sqrt{2}}{3} c m . A n s - O p t i o n D .$

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