# Geo4.38

Find the locus of point of intersection of two perpendicular tangents to the circle $x^{2}+y^{2}=a^{2}\,$

Given circle is

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

Let this equation be 1.

Let (x1,y1) be the point of intersection of two pependicular tangents.

Equation of a tangent to equation 1 is

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

Let this equation be 2.

The equation 2 passes thro' (x1,y1), we have

$y_{1}=mx_{1}+a{\sqrt {(1+m^{2})}}\,$

Simplifying

$(y_{1}-mx_{1})^{2}=a^{2}(1+m^{2})\,$

Rearranging

$(x_{1}-a^{2})m^{2}-2x_{1}y_{1}m+(y_{1}^{{2}}-a^{2})=0\,$

If m1,m2 are the two slopes of two tangents from (x1,y1) to 1,m1,m2 are the roots of the above quadratic equation,

Given two tangents are pependicular to each other, m1m2=-1

${\frac {y_{1}^{{2}}-a^{2}}{x_{1}^{{2}}-a^{2}}}=-1\,$

Hence

$x_{1}^{{2}}+y_{1}^{{2}}=2a^{2}\,$

Now the locus of the point (x1,y1) is

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