question_answer

From any point on the circle\[{{x}^{2}}+{{y}^{2}}={{a}^{2}},\]two tangents   are   drawn   to   the   circle \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}si{{n}^{2}}\alpha ,\]then angle between them is

A)  \[\frac{\alpha }{2}\]

B)  \[\alpha \]

C)  \[2\alpha \]              

D)  None of these

Correct Answer: C

Solution :

 Let PQ and PR are two tangents and point P is on the circle\[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\]with coordinates \[(a\cos t,a\sin t)\]and\[\angle OPQ=\theta \] Now,\[PQ=\]length of tangent drawn from the point P to the circle\[{{x}^{2}}+{{y}^{2}}={{a}^{2}}si{{n}^{2}}\alpha \] \[\therefore \]\[PQ=\sqrt{{{a}^{2}}{{\cos }^{2}}t+{{a}^{2}}{{\sin }^{2}}t-{{a}^{2}}{{\sin }^{2}}\alpha }\] \[=\sqrt{{{a}^{2}}-{{a}^{2}}{{\sin }^{2}}\alpha }\] \[=a\cos \alpha \] and \[OQ=\]radius of the circle \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}si{{n}^{2}}\alpha \] \[\Rightarrow \] \[OQ=a\text{ }sin\,\alpha \] \[\therefore \] \[\tan \theta =\frac{OQ}{QP}=\frac{a\sin \alpha }{a\cos \alpha }=\tan \alpha \] \[\theta =\alpha \] \[\therefore \]Angle between both the tangents\[=2\alpha \]