Question:
Let $P Q$ be a diameter of the circle $x^{2}+y^{2}=9$. If $\alpha$ and $\beta$ are the lengths of the perpendiculars from $P$ and $Q$ on the straight line, $x+y=2$ respectively, then the maximum value of $\alpha \beta$ is
Solution:
Let $\mathrm{P}(3 \cos \theta, 3 \sin \theta)$
$Q(-3 \cos \theta,-3 \sin \theta)$
$\Rightarrow \alpha \beta=\frac{\left|(3 \cos \theta+3 \sin \theta)^{2}-4\right|}{2}$
$\Rightarrow \alpha \beta=\frac{5+9 \sin 2 \theta}{2} \leq 7$