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The polar of P w.r.t the circle x^{2}+y^{2}=a^{2}\, touches the circle (x-f)^{2}+(y-g)^{2}=b^{2}\,.Prove that its locus is given by the equation (fx+gy-a)^{2}=b^{2}(x^{2}+y^{2})\,.

Let P be the point (x_{1},y_{1})\,

Polar of P w.r.t the circle is


This line touches the circle (x-f)^{2}+(y-g)^{2}=b^{2}\,

Hence,the length of the perpendicular from the center is equal to the radius.

centre of the circle is (f,g)\, and radius is b.


{\frac  |{fx_{1}+gy_{1}-a^{2}}}|{{\sqrt  {x_{1}^{{2}}+y_{1}^{{2}}}}}=b\,

|{fx_{1}+gy_{1}-a^{2}}|=b({\sqrt  {x_{1}^{{2}}+y_{1}^{{2}}}})\,

Squaring on both sides,


Hence the required locus is


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