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 \[=\,\angle \,QPR=2\alpha .\]You need to login to perform this action.
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