In the figure, $A$ and $B$ are centres of two circles touching each other at $M$ . Line $A C$ and line $B D$ are tangents. If $A D = 6 c m$ and $B C = 9 c m$ then the lengths of $s e g A C$ and $s e g B D$ are respectively

3 \sqrt{21}

12

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12, 3 \sqrt{21}

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15, 3 \sqrt{21}

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3 \sqrt{21}

15

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Solution

The correct option is B $12, 3 \sqrt{21}$
$G i v e n - P & Q a r e t h e c e t r e s o f t w o c i r c l e s o f r a d i u s 6 c m & 9 c m r e s p e c t i v e l y, t o u c h i n g e a c h o t h e r a t M . A C i s a t a n g e n t t o t h e b i g g e r c i r c l e a t C a n d B D i s a t a n g e n t t o t h e s m a l l e r c i r c l e a t D . T o f i n d o u t - t h e r e s p e c t i v e l e n g t h s o f s e g A C & s e g B D = ? S o l u t i o n - A M = A D = 6 c m (r a d i i o f t h e s a m e c i r c l e) & B M = B C = 9 c m . (r a d i i o f t h e s a m e c i r c l e) . W e k n o w t h a t t h e l i n e, j o i n i n g t h e c e n t r e s o f t w o c i r c l e s w h o t o u c h e a c h o t h e r, p a s s e s t h r o u g h t h e p o i n t o f c o n t a c t o f t h e c i r c l e s . ∴ B o t h A M & B M l i e o n t h e s a m e l i n e A B . S o A B = A M + B M = (6 + 9) c m = 15 c m . A g a i n w e k n o w t h a t i f a l i n e t o u c h e s a c i r c l e a t a p o i n t t h e n t h e r a d i u s t h r o u g h t h a t p o i n t i s p e r p e n d i c u l a r t o t h e t a n g e n t a t t h a t p o i n t . ∴ A D ⊥ B D & B C ⊥ A C . S o Δ A C B i s a r i g h t o n e w i t h A B a s h y p o t e n u s e . ∴ A p p l y i n g P y t h a g o r a s T h e o r e m, w e g e t A C = \sqrt{{A B}^{2} - {B C}^{2}} = \sqrt{15^{2} - 9^{2}} c m = 12 c m . S i m i l a r l y Δ A D B i s a r i g h t o n e w i t h A B a s h y p o t e n u s e . ∴ A p p l y i n g P y t h a g o r a s T h e o r e m, w e g e t B D = \sqrt{{A B}^{2} - {A D}^{2}} = \sqrt{15^{2} - 6^{2}} c m = 3 \sqrt{21} c m . ∴ s e g A C = 12 c m a n d s e g B D = 3 \sqrt{21} c m . A n s - O p t i o n B .$

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