Locus of point of intersection of tangents to the circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2$ which makes complimentary angles with x-axis is

$\begin{array}{l}Equation of \tan gent to the circle x^2+y^2=a^2 is\\ y=mx\pm a\sqrt{1+m^2}\Rightarrow {\left(y-mx\right)}^2=a^2\left(1+m^2\right)\\ {\mathrm{m}}^2\left({\mathrm{x}}^2-{\mathrm{a}}^2\right)-2\mathrm{mxy}+\left({\mathrm{y}}^2-{\mathrm{a}}^2\right)=0\\ {\mathrm{\theta }}_1+{\mathrm{\theta }}_2=\frac{\mathrm{\pi }}{2}\Rightarrow {\mathrm{tan\theta }}_1{\mathrm{tan\theta }}_2=1\\ {\mathrm{m}}_1{\mathrm{m}}_2=1\Rightarrow \frac{{\mathrm{y}}_1^2-{\mathrm{a}}^2}{{\mathrm{x}}_1^2-{\mathrm{a}}^2}=1\\ \Rightarrow {\mathrm{x}}_1^2-{\mathrm{y}}_1^2=0\end{array}$