Write True or False and justify your answer in each of the following :

If angle between two tangents drawn from a point P to a circle of radius a and centre O is $90^{0}$ , then OP $= a \sqrt{2}$ .

True

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False

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Ambiguous

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Data insufficient

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Solution

The correct option is D True
$G i v e n - P A & P B a r e t a n g e n t s t o t h e c i r c l e w i t h c e n t r e O a t A & B . ∠ O A P = 90^{o} T o f i n d o u t - T h e s t a t e m e n t, O P = a \sqrt{2}, i s t r u e o r f a l s e . J u s t i f i c a t i o n - P A & P B a r e t a n g e n t s t o t h e c i r c l e a t A & B . ∴ ∠ P A O & ∠ P B O a r e = 90^{o} e a c h . ∴ ∠ P A O + ∠ P B O = 2 \times 90^{o} = 180^{o} . . . . . . . (i) N o w i n q u a d r i l a t e r a l P A O B ∠ A P B + ∠ A O B + (∠ P A O + ∠ P B O) = 360^{o} (a n g l e s u m p r o p e r t y o f q u a d r i l a t e r a l s) ⟹ 90^{o} + ∠ A O B + 180^{o} = 360^{o} (f r o m i) ⟹ ∠ A O B = 90^{o} . A n d O A = O B (r a d i i o f t h e s a m e c i r c l e) ∴ O A P B i s a s q u a r e w i t h O P a s i t s d i a g o n a l & s i d e = a . ∴ O P = a \sqrt{2} A n s - T r u e .$

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