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  • From a point on the circle , two tangents are drawn to the circle <img alt="x^2+y^2=a^2 \

From a point on the circle x^2+y^2=a^2, two tangents are drawn to the circle x^2+y^2=a^2 \sin ^2 \alpha. The angle between them is

Option: 1

\alpha


Option: 2

\frac{\alpha}{2}


Option: 3

2 \alpha


Option: 4

None of these


Answers (1)

Let any point on the circle x^2+y^2=a^2 be (a \operatorname{cost}, a \sin t) and  \angle O P Q=\theta

Now; P Q= length of tangent from P on the circle x^2+y^2=a^2 \sin ^2 \alpha

\begin{aligned} \therefore \quad P Q & =\sqrt{a^2 \cos ^2 t+a^2 \sin ^2 t-a^2 \sin ^2 \alpha}=a \cos \alpha \\\\ O Q & =\text { Radius of the circle } x^2+y^2=a^2 \sin ^2 \alpha \end{aligned}

O Q=a \sin \alpha \quad \therefore \quad \tan \theta=\frac{O Q}{P Q}=\tan \alpha \Rightarrow \theta=\alpha ; \quad \therefore \quad \text { Angle between tangents } \angle Q P R=2 \alpha \text {. }
Alternative Method: We know that, angle between the tangent from (\alpha, \beta) to the circle 2 \tan ^{-1}\left(\frac{a}{\sqrt{\alpha^2+\beta^2-a^2}}\right).Let point on the circle x^2+y^2=a^2 be (a \cos t, a \sin t)
Angle between tangent =
2 \tan ^{-1}\left(\frac{a \sin \alpha}{\sqrt{a^2 \cos ^2 t+a^2 \sin ^2 t-a^2 \sin ^2 \alpha}}\right)=2 \tan ^{-1}\left(\frac{a \sin \alpha}{a \cos \alpha}\right)=2 \alpha

Posted by

Ramraj Saini

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