Show that the points of intersection of the perpendicular tangents to an Ellipse lies on a circle

Let the equation of the ellipse be $\mathrm{S}=\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}-1=0$
Let P(x₁, y₁) be the point of intersection of two perpendicular tangents drawn to the ellipse.
Let $\mathrm{y}=\mathrm{mx}\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2}$ be a tangent to the ellipse S = 0 passing through P.
Then ${\mathrm{y}}_1={\mathrm{mx}}_1\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2}$
$\begin{array}{l}\Rightarrow {\mathrm{y}}_1={\mathrm{mx}}_1\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2}\\ \Rightarrow {\mathrm{y}}_1-{\mathrm{mx}}_1=\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2}\\ \Rightarrow {\left({\mathrm{y}}_1-{\mathrm{mx}}_1\right)}^2={\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2\\ \Rightarrow {\mathrm{y}}_1^2+{\mathrm{m}}^2{\mathrm{x}}_1^2-2{\mathrm{ymx}}_1={\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2 \\ \Rightarrow \left({\mathrm{x}}_1^2-{\mathrm{a}}^2\right){\mathrm{m}}^2-2{\mathrm{x}}_1{\mathrm{y}}_1\mathrm{m}+\left({\mathrm{y}}_1^2-{\mathrm{b}}^2\right)=0 ...\left(1\right)\end{array}$
Let m₁, m₂ are the slopes of the tangents through P and m₁, m₂ are the roots of (1).
The tangents through P are perpendicular
$\begin{array}{l}\Rightarrow {\mathrm{m}}_1{\mathrm{m}}_2=-1\Rightarrow \frac{{\mathrm{y}}_1^2-{\mathrm{b}}^2}{{\mathrm{x}}_1^2-{\mathrm{a}}^2}=-1\\ \Rightarrow {\mathrm{y}}_1^2-{\mathrm{b}}^2={\mathrm{x}}_1^2+{\mathrm{a}}^2\Rightarrow {\mathrm{x}}_1^2+{\mathrm{y}}_1^2={\mathrm{a}}^2+{\mathrm{b}}^2\end{array}$
$\therefore \mathrm{P}\text{ lies on }{\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2+{\mathrm{b}}^2$ which is a circle with centre at origin, which is the centre of the ellipse.