Q.Tangents to the circle $x^2+y^2=a^2$ with inclinations $\theta_1$ and $\theta_2$ intersect at $P$. Find the locus of $P$ such that $\cot\theta_1+\cot\theta_2=5$.

The distance from $P(6,8)$ to the centre $O(0,0)$ is $OP=\sqrt{6^2+8^2}=\sqrt{100}=10$.

For a tangent from $P$ touching the circle of radius $r$ at $A$, triangle $OAP$ is right-angled at $A$ (radius $\perp$ tangent), so the tangent length is $PA=PB=\sqrt{OP^2-r^2}=\sqrt{100-r^2}$.

The quadrilateral $PAOB$ consists of two congruent right triangles $OAP$ and $OBP$, so

$\text{Area}=2\cdot\frac12\,(OA)(PA)=r\sqrt{100-r^2}.$ …