Two circles of radii $8 c m$ and $3 c m$ have their centres $13 c m$ apart. The length of a direct common tangent to the two circles is

12 c m

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24 c m

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18 c m

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13 c m

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Solution

The correct option is A $12 c m$
$A B I s t h e c o m m o n t a n g e n t o f t w o c i r c l e s, w h o t o u c h e a c h o t h e r . T h e c e n t r e o f o n e c i r c l e i s P w i t h r a d i u s = 3 c m a n d t h e c e n t r e o f a n o t h e r c i r c l e i s Q w i t h r a d i u s = 8 c m . S o l u t i o n - W e j o i n P Q, A P & P Q . A p e r p e n d i c u l a r P N i s d r o p p e d f r o m P t o B Q a t N . N o w (A P & B Q) ⊥ A B s i n c e t h e r a d i u s t h r o u g h t h e p o i n t o f c o n t a c t o f a t a n g e n t t o a c i r c l e i s p e r p e n d i c u l a r t o t h e t a n g e n t . ∴ ∠ P A B = 90^{o} = ∠ Q B A . A l s o ∠ P N B = {∠ P N Q = 90}^{o} s i n c e P N ⊥ P Q . ∴ A B N P i s a r e c t a n g l e . S o B N = A P = 3 c m a n d P N = A B . ∴ N Q = B Q - B N = (8 - 3) c m = 5 c m . A g a i n P Q = 13 c m . ∴ Δ P Q B i s a r i g h t t r i a n g l e w i t h P Q a s h y p o t e n u s e . ∴ B y P y t h a g o r a s t h e o r e m, w e g e t P N = \sqrt{{P Q}^{2} - {N Q}^{2}} = \sqrt{13^{2} - 5^{2}} c m = 12 c m . ∴ P N = A B = 12 c m . A n s - O p t i o n B .$

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