Two circles touch externally at $P$ and a common tangent touches them at $A a n d B$ . Prove that the common tangent at $P$ bisects $A B$ and $A B$ subtends a right angle at $P$ .

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Solution

$G i v e n : P A T i s a t a n g e n t t o t h e c i r c u m c i r c l e o f a △ A B C a t t h e v e r t e x A . A l i n e p a r a l l e l t o P A T i n t e r s e c t s t h e s i d e s A B a n d A C a t t h e p o i n t s D a n d E r e s p e c t i v e l y . T o P r o v e : T h e c o m m o n t a n g e n t a t P b i s e c t s A B . ∠ A P B i s a r i g h t a n g l e P r o o f : L e t P T b e t h e c o m m o n t a n g e n t a t a n y p o i n t P . S i n c e t h e t a n g e n t t o a c i r c l e f r o m a n e x t e r n a l p o i n t a r e e q u a l, ∴ T A = T P, T B = T P ∴ T A = T B i . e ., P T b i s e c t s A B a t T I n △ P T A T A = T P \Rightarrow ∠ T A P = ∠ T P A I n △ P B T T B = T P g i v e s ∠ T B P = ∠ T P B ∴ ∠ T A P + ∠ T B P = ∠ T P A + ∠ T P B = A P B \Rightarrow ∠ T A P + ∠ T B P + ∠ A P B = 2 ∠ A P B \Rightarrow 2 ∠ A P B = 180 [s u m o f a n g l e s o f a △ = 180] \Rightarrow ∠ A P B = 90 H e n c e p r o v e d$ .

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Similar questions

In the given figure, two circles touch each other externally at point $P$ . $A B$ is the direct common tangent of these circles. Prove that :

(i) tangent at point $P$ bisects $A B$ ,

(ii) angle $A P B =$ $90^{\circ}$ .

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APB is

(a) 30º (b) 45º (c) 60º (d) 90º [CBSE 2014]

Q. In the given figure, two circles touch each other at the point

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Q

In the given figure, PT is a common tangent to the circles touching externally at P and AB is another common tangent touching the circles at A and B. Prove that: [3 MARKS]

(i) T is the mid-point of AB

$(i i) ∠ A P B = 90^{\circ}$

(iii) If X and Y are centers of the two circles, show that the circle on AB as diameter touches the line XY.

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Two circles touch externally at P and a common tangent touches them at AandB. Prove that the common tangent at P bisects AB and AB subtends a right angle at P.

Two circles touch externally at $P$ and a common tangent touches them at $A a n d B$ . Prove that the common tangent at $P$ bisects $A B$ and $A B$ subtends a right angle at $P$ .