# Geo5.2.40

Show that the equation of the auxiliary circle of the ellipse ${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1\,$ is $x^{2}+y^{2}=a^{2}\,$.

The equation to the ellipse is ${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1\,$

The equation to the tangent of the ellipse is $y=mx\pm {\sqrt {a^{2}m^{2}+b^{2}}},y-mx=\pm {\sqrt {a^{2}m^{2}+b^{2}}}\,$

The equation to the perpendicular from either focus $(\pm ae,0)\,$ on this tangent is given by $y={\frac {-1}{m}}(x\pm ae),my+x=\pm ae\,$

To find the locus of point of intersection of the above two equations,eliminate m from them.

Squaring and adding the two equations,we get

$(y-mx)^{2}+(my+x)^{2}=a^{2}m^{2}+b^{2}+a^{2}e^{2}\,$

$x^{2}(1+m^{2})+y^{2}(1+m^{2})=a^{2}m^{2}+a^{2}(1-e^{2})+a^{2}e^{2}\,$

$(x^{2}+y^{2})(1+m^{2})=a^{2}(m^{2}+1)\,$

$x^{2}+y^{2}=a^{2}\,$

Therefore the locus is the auxiliary circle concentric with the ellipse.