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From any point on the circle ${x^2} + {y^2} = {a^2}$ tangents are drawn to the circle ${x^2} + {y^2} = {a^2}{\sin ^2}\alpha $, the angle between them is
$\frac{\alpha }{2}$
$\alpha $
$2\alpha $
None of these
Solution
(c) Let any point on the circle
${x^2} + {y^2} = {a^2}$ be $(a\cos t,\,\,a\sin t)$ and $\angle \,OPQ = \theta $
Now; $PQ = $ length of tangent from $P$ on the circle ${x^2} + {y^2} = {a^2}{\sin ^2}\alpha $
$\therefore $ $PQ = $$\sqrt {{a^2}{{\cos }^2}t + {a^2}{{\sin }^2}t – {a^2}{{\sin }^2}\alpha } $$ = a\cos \alpha $
$OQ = $ Radius of the circle
${x^2} + {y^2} = {a^2}{\sin ^2}\alpha $
$OQ = $ $a\sin \alpha $,
$\therefore $$\tan \theta = \frac{{OQ}}{{PQ}} = \tan \alpha $
$\Rightarrow \,\theta = \alpha $
$\therefore $ Angle between tangents $ = \,\angle \,QPR = 2\alpha .$
