If ${\mathrm{\theta }}_1\text{ and }{\mathrm{\theta }}_2$ are the angles of inclination of tangents through a point P to the circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2$, then find the locus of P when $\cot {\mathrm{\theta }}_1+\cot {\mathrm{\theta }}_2=\mathrm{k}$ 

The equation of the tangent to ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2$ having slope m is $\mathrm{y}=\mathrm{mx}\pm \mathrm{a}\sqrt{1+{\mathrm{m}}^2} ...\left(1\right)$
Let P(x₁, y₁) be a point on the locus
If (1) passes through P, then
$\begin{array}{l}{\mathrm{y}}_1={\mathrm{mx}}_1\pm \mathrm{a}\sqrt{1+{\mathrm{m}}^2}\\ \Rightarrow {\left({\mathrm{y}}_1-{\mathrm{mx}}_1\right)}^2={\mathrm{a}}^2\left(1+{\mathrm{m}}^2\right)\\ \Rightarrow \left({\mathrm{x}}_1^2-{\mathrm{a}}^2\right){\mathrm{m}}^2-2{\mathrm{x}}_1{\mathrm{y}}_1\mathrm{m}+\left({\mathrm{y}}_1^2-{\mathrm{a}}^2\right)=0\end{array}$
If m₁, m₂ are the roots of the above equation, then
$\begin{array}{l}{\mathrm{m}}_1+{\mathrm{m}}_2=\tan {\mathrm{\theta }}_1+\tan {\mathrm{\theta }}_2=\frac{2{\mathrm{x}}_1{\mathrm{y}}_1}{{\mathrm{x}}_1^2-{\mathrm{a}}^2}\\ {\mathrm{m}}_1{\mathrm{m}}_2=\tan {\mathrm{\theta }}_1\tan {\mathrm{\theta }}_2=\frac{{\mathrm{y}}_1^2-{\mathrm{a}}^2}{{\mathrm{x}}_1^2-{\mathrm{a}}^2}\end{array}$
$\text{ Given }\cot {\mathrm{\theta }}_1+\cot {\mathrm{\theta }}_2=\mathrm{k}$
$\begin{array}{l}\Rightarrow \tan {\mathrm{\theta }}_1+\tan {\mathrm{\theta }}_2=\mathrm{k}\left(\tan {\mathrm{\theta }}_1\cdot \tan {\mathrm{\theta }}_2\right)\\ \Rightarrow \frac{2{\mathrm{x}}_1{\mathrm{y}}_1}{{\mathrm{x}}_1^2-{\mathrm{a}}^2}=\mathrm{k}\left(\frac{{\mathrm{y}}_1^2-{\mathrm{a}}^2}{{\mathrm{x}}_1^2-{\mathrm{a}}^2}\right)\end{array}$
$\therefore \text{ The locus of }\mathrm{P}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\text{ is }\mathrm{k}\left({\mathrm{y}}^2-{\mathrm{a}}^2\right)=2\mathrm{xy}$