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 $\tan\theta_1+\tan\theta_2=0$.

For tangents from $P(x,y)$ to $x^2+y^2=a^2$ with slopes $m_1=\tan\theta_1$, $m_2=\tan\theta_2$, the governing quadratic $(x^2-a^2)m^2-2xym+(y^2-a^2)=0$ gives $m_1+m_2=\dfrac{2xy}{x^2-a^2}$. Setting $\tan\theta_1+\tan\theta_2=m_1+m_2=0$: …

Apply the sum-of-slopes formula from the two-tangent quadratic, set it …

• Forgetting the denominator $x^2-a^2$ must be nonzero for this division to be valid (points where $x^2=a^2$ need separate consideration) …