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Area of a Circle

Two circles t...

Question

Two circles touch internally. The sum of their areas is $116 π$ sq. cm and the distance between their centers is 6 cm. The radii of the given circles are

A

8cm & 20cm.

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B

4cm & 10cm.

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C

6cm & 8cm

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D

5cm & 9cm.

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Solution

The correct option is B 4cm & 10cm.
$G i v e n - T w o c i r c l e s t o u c h i n t e r n a l l y . T h e d i s t a n c e b e t w e e n t h e i r c e r t r e s = 6 c m . T h e r a d i i o f t h e o u t e r & i n n e r c i r c l e s a r e r_{2} & r_{1} r e s p e c t i v e l y . T h e s u m o f t h e i r a r e a s = 116 π .$

$T o f i n d o u t - {(r}_{2} & r_{1}) = ?$
$S o l u t i o n - S i n c e t h e c i r c l e s t o u c h i n t e r n a l l y, t h e d i s t a n c e b e t w e e n t h e i r c e r t r e s = t h e d i f f e r e n c e c e b e t w e e n t h e i r r a d i i = 6 c m .$

$∴ r_{2} - r_{1} = 6 c m ⟹ r_{2} = r_{1} + 6. S o a r . o u t e r c i r c l e = π \times {(r_{2})}^{2} = π \times {(r_{1} + 6)}^{2} . A n d a r . i n n e r c i r c l e = π \times {(r_{1})}^{2} .$

$∴ T h e s u m o f t h e i r a r e a s = π \times {(r_{1} + 6)}^{2} + π \times {(r_{1})}^{2} = 116 π . ⟹ {(r_{1})}^{2} + 6 r - 40 = 0 ⟹ (r_{1} + 10) (r_{1} - 4) = 0 ⟹ r_{1} = (- 10, 4) c m .$
$W e r e j e c t t h e n e g a t i v e v a l u e o f r_{1} s i n c e r_{1} i s a l e n g t h . ∴ r_{1} = 4 c m . S o r_{2} = (r_{1} + 6) c m = (4 + 6) c m = 10 c m . ∴ r_{1} = 4 c m, r_{2} = 10 c m .$

$A n s - O p t i o n B .$

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