# Let P Q be a diameter of the circle

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 $P(3 \cos \theta, 3 \sin \theta), Q(-3 \cos \theta,-3 \sin \theta)$

$\alpha=\left|\frac{3 \cos \theta+3 \sin \theta-2}{\sqrt{2}}\right|, \beta=\left|\frac{-3 \cos \theta-3 \sin \theta-2}{\sqrt{2}}\right|$

$\alpha \beta=\left|\frac{(3 \cos \theta+3 \sin \theta)^{2}-4}{2}\right|=\left|\frac{5+9 \sin 2 \theta}{2}\right|$

$\alpha \beta$ is man. when $\sin 2 \theta=1$

$\left.\therefore \alpha \beta\right|_{\max }=\frac{5+9}{2}=7$