question_answer

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 [RPET 2002]

A)            \[\frac{\alpha }{2}\]                

B)            \[\alpha \]

C)            \[2\alpha \]                               

D)            None of these

Correct Answer: C

Solution :

           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 \[=\,\angle \,QPR=2\alpha .\]