Two circles of radii $8 c m$ and $3 c m$ respectively touch internally then show that the distance between their centres is equal to the difference of their radii, and that distance is

13cm

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8cm

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5cm

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3cm

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Solution

The correct option is C 5cm
$G i v e n - T h e c i r c l e s, C_{1} w t t h c e n t r e O & o f r a d i u s r_{1} = 8 c m a n d C_{2} w t t h c e n t r e P & o f r a d i u s r_{2} = 3 c m, t o u c h e a c h o t h e r i n t e r n a l l y a t Q . (i) T o s h o w t h a t - O P = r_{1} - r_{2} a n d (i i) O P = r_{1} - r_{2} = ? S o l u t i o n - W e k n o w t h a t i f t w o c i r c l e s t o u c h i n t e r n a l l y a t a p o i n t t h e n t h e p o i n t o f c o n t a c t a n d t h e c e n t r e s o f t h e c i r c l e s l i e o n t h e s a m e l i n e w h i c h c o n t a i n s t h e d i a m e t e r s o f t h e c i r c l e s . (i) H e r e t h e p o i n t o f c o n t a c t i s Q, t h e c e n t r e o f C_{1} i s O a n d t h a t o f C_{2} i s P . ∴ T h e l i n e S Q c o n t a i n s t h e d i a m e t e r s o f b o t h C_{1} & C_{2} . i . e S, O, P a n d Q l i e o n t h e s a m e l i n e . S o T h e r a d i u s o f C_{1} = r_{1} i s O Q a n d t h e r a d i u s o f C_{1} = r_{1} i s P Q o r P R . ∴ T h e d i s t a n c e b e t w e e n t h e c e n t r e s o f C_{1} & C_{2} i s O P = O Q - P Q = r_{1} - r_{2} (i i) O P = O Q - P Q = r_{1} - r_{2} = 8 c m - 3 c m = 5 c m . A n s - O p t i o n C .$

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