The tangents to $x^2+y^2=a^2$ having inclinations $\alpha$ and $\beta$ intersect at $P.$If $\cot \alpha +\cot \beta =0$ then the locus of $P$ is 

Let the coordinates of$P$ be$(h, k).$

Let the equation of a tangent from $P(h, k)$ to the circle

$x^2+y^2=a^2$ be $y=mx+a\sqrt{1+m^2}$

Since $P(h, k)$ lies on $y=mx+a\sqrt{1+m^2}.$

$\begin{array}{l}\therefore k=mh+a\sqrt{1+m^2}\\ \Rightarrow (k-mh{)}^2=a\left(1+m^2\right)\\ \Rightarrow m^2\left(h^2-a^2\right)-2mkh+k^2-a^2=0\end{array}$

This is a quadratic in m. Let the two roots bem1 and $m_2$.Then,

$m_1+m_2=\frac{2hk}{h^2-a^2}$

But, tan $a= m_1$, $\tan \beta =m_2$ and it is given that 

$\cot \alpha +\cot \beta =0$

$\Rightarrow \frac{1}{m_1}+\frac{1}{m_2}=0\Rightarrow m_1+m_2=0\Rightarrow \frac{2hk}{k^2-a^2}=0\Rightarrow hk=0$

Hence, the locus of $(h, k)$ is $xy = 0.$