Geo4.54

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Show that the locus of the poles of the tangents to the circle x^{2}+y^{2}=a^{2}\, with respect to the circle (x+a)^{2}+y^{2}=2a^{2}\, is y^{2}+4ax=0\,

Let (x1,y1) be a pole w.r.t the circle (x+a)^{2}+y^{2}=2a^{2}\,

x^{2}+y^{2}+2ax-a^{2}=0\,

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

x(x_{1}+a)+yy_{1}+(ax_{1}-a^{2})=0\,

This is a tangent to the equation x^{2}+y^{2}=a^{2}\,

Hence

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

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

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

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

y_{1}^{{2}}=-4ax_{1}\,

Now the locus of the poles of the tangent is

y^{2}+4ax=0\,

Main Page:Geometry:Circles