Question

Find the locus of the point of intersection of two perpendicular tangents to a given circle $x^2+y^2=a^2$?

Answer 1

The equation of tangent to the circle $x^2+y^2=a^2$ is

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

$P(h,k)$ lies on tangent then

$k-mh=a\sqrt{1+m^2}$

$\Rightarrow {(k-mh)}^2={\left(a\sqrt{1+m^2}\right)}^2$

$\Rightarrow {(k-mh)}^2=a^2(1+m^2)$

$\Rightarrow k^2+m^2{h}^2-2kmh=a^2+a^2{m}^2$

$\Rightarrow k^2+m^2{h}^2-2kmh-a^2-a^2{m}^2=0$

$\Rightarrow m^2(h^2-a^2)-2mhk+(k^2-a^2)=0$

This is the quadratic equation in $m$

Let the roots be $m_1$ and $m_2$ then

$m_1{m}_2=-1$ since the tangents are perpendicular to each other.

$\frac{k^2-a^2}{h^2-a^2}=-1$

$\Rightarrow k^2-a^2=-h^2+a^2$

$\Rightarrow k^2+h^2=2a^2$

Hence locus $P(h,k)$ is $x^2+y^2=2a^2$