Two circles of unit radius touch each other and each of them touches internally a circle of radius 2 units, as shown in the following figure. The radius of the circle which touches all the three circles is

5

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\frac{3}{2}

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\frac{2}{3}

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None of these

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Solution

The correct option is C $\frac{2}{3}$
$G i v e n - C & A a r e c e n t r e s o f t w o c i r c l e s o f r a d i i = 1 u n i t a n d t h e y t o u c h a t B . T h e s e t w o c i r c l e s i n t e r n a l l y t o u c h a n o t h e r c i r c l e w h o s e r a d i u s i s 2 u n i t s i . e d i a m e t e r = 2 \times 2 u n i t s = 4 u n i t s . . T h e r e i s a t h i r d c i r c l e w i t h c e n t r e a s O a n d i t t o u c h e s a l l t h e t h r e e p r e v i o u s c i r c l e s . T o f i n d o u t - t h e r a d i u s = r o f t h e c i r c l e w i t h c e n t r e O . S o l u t i o n - B D = 2 A D = 2 \times 1 u n i t = 2 u n i t s . N o w B E & B D a r e s a m e s t r a i g h t l i n e s i n c e w h e n t w o c i r c l e s t o u c h e a c h o t h e r t h e n t h e l i n e j o i n i n g t h e c e n t r e s w i l l p a s s t h r o u g h t h e p o i n t o f c o n t a c t . B u t E D = B E + A D = (2 + 2) u n i t s = 4 u n i t s w h i c h i s t h e d i a m e t e r o f t h e e n c l o s i n g c i r c l e . N o w B T i s t h e c o m m o n t a n g e t o f t h e c i r c l e s w i t h c e n t r e s A & C . ∴ B T ⊥ E D . i . e Δ O A B i s a r i g h t o n e w i t h h y p o t e n u s e a s O A = 1 + r . (O A w i l l p a s s t h r o u g h P s i n c e t h e s e c i r c l e s t o u c h a t P) A l s o B T i s t h e r a d i u s o f t h e e n c l o s i n g c i r c l e . i . e B T = 2 u n i t s a n d O B = B T - O T = 2 - r . S o, a p p l y i n g P y t h a g o r a s t h e o r e m i n Δ O A B, w e h a v e O A^{2} = {O B}^{2} + {A B}^{2} ⟹ {(1 + r)}^{2} = {(2 - r)}^{2} + {(1)}^{2} ⟹ 4 - 4 r = 2 r ⟹ r = \frac{2}{3} u n i t . A n s - O p t i o n C .$

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