If from any point on the circle $x^{2} + y^{2} = a^{2}$ , tangents are drawn to the circle $x^{2} + y^{2} = b^{2}$ $(a > b)$ , then the angle between the tangents is $k {sin}^{- 1} \frac{b}{a}$ , where $k =$

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Solution

The correct option is A $2$
Let $P$ be any point on $x^{2} + y^{2} = a^{2}$
Let $P T, P T^{'}$ be the tangents to the concentric circle $x^{2} + y^{2} = b^{2}$
Join $O P, O T$ and $O T$
Then $∠ O P T = ∠ O P T = θ$ (say)
and $O T ⊥ P T$
Also, $O T = b$ and $O P = a$
From right angled triangle $O P T,$ $sin θ = \frac{b}{a}$ $\Rightarrow θ = {sin}^{- 1} \frac{b}{a}$
$∴$ angles between the tangents $= 2 θ = 2 {sin}^{- 1} \frac{b}{a},$
$∴ k = 2$

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