Question

Let $PQ$ 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

Answer 1

$\alpha =\left|\frac{3\cos \theta +3\sin \theta +2}{\sqrt{2}}\right|$
$\beta =\left|\frac{3\cos \theta +3\sin \theta -2}{\sqrt{2}}\right|$
Now,
$\alpha \beta =\left|\frac{(3\cos \theta +3\sin \theta {)}^2-4}{2}\right|$
$\Rightarrow \alpha \beta =\left|\frac{9+9\sin 2\theta -4}{2}\right|$
$\Rightarrow \alpha \beta =\left|\frac{5+9\sin 2\theta }{2}\right|$
$\therefore \alpha {\beta }_{max}=\frac{9+5}{2}=7$