Geo4.72

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If the polar of the point (x_{1},y_{1})\, w.r.t the circle x^{2}+y^{2}=a^{2}\, touches the circle (x-a)^{2}+y^{2}=a^{2}\,, show that the point lies on the curve y^{2}+2ax=a^{2}\,

The polar of (x_{1},y_{1})\, w.r.t the circle x^{2}+y^{2}=a^{2}\,

is

xx_{1}+yy_{1}-a^{2}=0\,

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

The length of the perpendicular from the center (a,0) to the line is equal to radius.

Hence

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

(x_{1}-a)^{2}=(x_{1}^{{2}}+y_{1}^{{2}})\,

y_{1}^{{2}}+2ax_{1}=a^{2}\,

Therefore (x_{1},y_{1})\, lies on the curve

y^{2}+2ax=a^{2}\,


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