In the figure, points $P$ and $Q$ are the centres of the circles. Radius $Q N = 3 c m$ , $P Q = 9 c m$ . $M$ is the point of contact of the circles. Line $N D$ is the tangent to the larger circle. Point $C$ lies on the smaller circle.The respective lengths of $N C$ , $N D$ and $C D$ are

N C = 3 \sqrt{3} c m, N D = 3 \sqrt{3} c m & C D = 6 \sqrt{3} c m .

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N C = 6 \sqrt{3} c m, N D = 6 \sqrt{3} c m & C D = 12 \sqrt{3} c m .

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N C = 3 \sqrt{3} c m, N D = 6 \sqrt{3} c m & C D = 3 \sqrt{3} c m .

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N C = 6 \sqrt{3} c m, N D = 12 \sqrt{3} c m & C D = 6 \sqrt{3} c m .

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Solution

The correct option is C $N C = 3 \sqrt{3} c m, N D = 6 \sqrt{3} c m & C D = 3 \sqrt{3} c m .$
$G i v e n - Q & P a r e t h e c e n t r e s o f t w o c i r c l e s C_{1} & C_{2} r e s p e c t i v e l y . T h e y t o u c h e a c h o t h e r e x t e r n a l l y a t M . T h e r a d i u s o f C_{1} i s Q N = 3 c m . T h e r a d i u s o f C_{2} i s P Q = 9 c m . N D i s a t a n g e n t t o C_{2} f r o m N a t D . N D i n t e r s e c t s C_{1} a t C . T o f i n d o u t - N C = ? N D = ? C D = ? S o l u t i o n - P o i n t s Q, M & P l i e o n t h e s a m e l i n e s i n c e t w o c i r c l e s t o u c h i n g e a c h o t h e r e x t e r n a l l y h a v e t h e i r c e n t r e s a n d t h e p o i n t o f c o n t a c t l i e o n t h e s a m e l i n e . B u t t h e l i n e N Q M i s a d i a m e t e r o f t h e c i r c l e a s p r e t h e g i v e n f i g u r e . ∴ N, Q, M & P l i e o n t h e s a m e s t r a i g h t l i n e . S o Q N = 3 c m = Q M, M P = P Q - Q M = (9 - 3) c m = 6 c m = P D . S o N M = (3 + 3) c m a n d N P = M N + M P = (6 + 6) c m = 12 c m . . . . . . . . . . . (i) W e j o i n C M & P D . I n Δ P N D w e h a v e ∠ N D P = 90^{o} (N D i s a t a n g e n t a t D a n d P D i s t h e r a d i u s t h r o u g h D .) ∴ Δ P N D i s a r i g h t o n e w i t h P N a s h y p o t e n u s e . S o, b y P y t h a g o r a s T h e o r e m, w e h a v e N D = \sqrt{{N P}^{2} - {P D}^{2}} = \sqrt{12^{2} - 6^{2}} c m = 6 \sqrt{3} c m . N o w ∠ N C M = 90^{o} (a n g l e i n a s e m i c i r c l e) . ∴ b e t w e e n Δ N C M & Δ N D P w e h a v e ∠ N C M = 90^{o} = ∠ N D P a n d ∠ P N D i s c o m m o n . S o Δ N C M & Δ N D P a r e s i m i l a r o n e s . ∴ \frac{N C}{N D} = \frac{N M}{N P} ⟹ N C = N D \times \frac{N M}{N P} = 6 \sqrt{3} \times \frac{6}{12} c m = 3 \sqrt{3} c m . A l s o C D = N D - N C = 6 \sqrt{3} - 3 \sqrt{3} c m = 3 \sqrt{3} c m . S o N C = 3 \sqrt{3} c m, N D = 6 \sqrt{3} c m & C D = 3 \sqrt{3} c m . A n s - O p t i o n C .$

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