question_answer

If from any point P on the circle \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\],  tangents are drawn to the circle \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c{{\sin }^{2}}a+({{g}^{2}}+{{f}^{2}}){{\cos }^{2}}a=0\] , then angle between the tangents is

A)  \[\alpha \]

B)  \[2\alpha \]

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

D)  None of these

Correct Answer: A

Solution :

Let \[p({{x}_{1}},{{y}_{1}})\] be any point on the circle \[{{x}^{2}}+{{y}^{2}}+2gx+2fy=c=0\] Then, the length of the tangents drawn from \[p({{x}_{1}},{{y}_{1}})\] to the circle \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c{{\sin }^{2}}\alpha \] \[+({{g}^{2}}+{{f}^{2}}){{\cos }^{2}}\alpha =0\]is \[PQ=PR\] \[=\sqrt{\left[ \begin{align}   & x_{1}^{2}+y_{1}^{2}+2g{{x}_{1}}+2f{{y}_{1}}+c{{\sin }^{2}}\alpha  \\  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+({{g}^{2}}+{{f}^{2}}){{\cos }^{2}}\alpha  \\ \end{align} \right]}\] \[=\sqrt{-c+c{{\sin }^{2}}\alpha +({{g}^{2}}+{{f}^{2}}){{\cos }^{2}}\alpha }\] \[=(\sqrt{{{g}^{2}}+{{f}^{2}}-c)}\cos \alpha \] The radius of the circle \[{{x}^{2}}+{{y}^{2}}+2gx+2fy=c{{\sin }^{2}}\alpha \] \[+({{g}^{2}}+{{f}^{2}}){{\cos }^{2}}\alpha =0\]is \[CQ=CR\] \[=\sqrt{{{g}^{2}}+{{f}^{2}}-c{{\sin }^{2}}\alpha -({{g}^{2}}+{{f}^{2}}){{\cos }^{2}}\alpha }\] \[=\sqrt{({{g}^{2}}+{{f}^{2}}-c}\sin \alpha \] In \[\Delta PQC,\]                            \[\tan \theta =\frac{CO}{PQ}\] \[\Rightarrow \]\[\tan \theta =\frac{\sqrt{({{g}^{2}}+{{f}^{2}}-c)}\sin \alpha }{\sqrt{({{g}^{2}}+{{f}^{2}}-c)}\cos \alpha }\]\[=\tan \alpha \] \[\Rightarrow \]\[\theta =\alpha \]