In the figure, $P Q$ is a transverse common tangent to two circles with centres $A$ and $B$ and of radii $5 c m$ and $3 c m$ respectively. If $P Q$ intersects $A B$ at $C$ such that $C P = 12 c m$ , then $A B$ is

20.8cm

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16.54cm

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42.84cm

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15.84cm

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Solution

The correct option is A 20.8cm
$G i v e n - P Q i s a n o b l i q u e t a n g e n t o f t w o c i r c l e s w i t h c e n t r e s A & B a n d r a d i i 5 c m & 3 c m r e s p e c t i v e l y . T h e p o i n t o f c o n t a c t o f P Q w i t h t h e c i r c l e w i t h c e n t r e A i s P a n d t h e p o i n t o f c o n t a c t o f P Q w i t h t h e c i r c l e w i t h c e n t r e B i s Q A B i n t e r s e c t s P Q a t C . C P = 12 c m . T o f i n d o u t - A B = ? S o l u t i o n - w e k n o w t h a t t h e r a d i u s t h r o u g h t h e p o i n t o f c o n t a c t o f a t a n g e n t t o a c i r c l e i s p e r p e n d i c u l a r t o t h e t a n g e n t . ∴ B o t h A P & B Q a r e p e r p e n d i c u l a r t o P Q . i . e ∠ A P C = ∠ B Q C = 90^{o} . A l s o ∠ A C P = ∠ B C Q (v e r t i c a l l y o p p o s i t e a n g l e s) . S o Δ A P C & Δ B Q C a r e s i m i l a r . ∴ \frac{C Q}{B Q} = \frac{C P}{P A} ⟹ C Q = \frac{B Q \times C P}{P A} = \frac{12 \times 3}{5} c m = 7.2 c m . N o w ∠ A P C = ∠ Q B C = 90^{o} i . e Δ A P C & Δ Q B C a r e r i g h t t r i a n g l e s w i t h A C & B C a s h y p o t e n u s e s r e s p e c t i v e l y . S o, a p p l y i n g P y t h a g o r a s t h e o r e m, w e g e t A C = \sqrt{{P A}^{2} + {P C}^{2}} = \sqrt{12^{2} + 5^{2}} c m = 13 c m . a n d B C = \sqrt{{C Q}^{2} + B Q^{2}} = \sqrt{{7.2}^{2} + 3^{2}} c m = 7.8 c m ∴ A B = A C + B C = (13 + 7.8) c m = 20.8 c m . A n s - O p t i o n A .$

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