Let PQ be a diameter of the circle x^2+y^2=9.

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: 7

Accepted Answer

$P=(3\cos\theta,3\sin\theta)$, $Q=-P$. Let $u=3\cos\theta+3\sin\theta\in[-3\sqrt2,3\sqrt2]$:
$$\alpha\beta=\frac{|u-2||u+2|}2=\frac{|u^2-4|}2.$$

Maximum at $u^2=18$: $\dfrac{|18-4|}2=7$.

Solutions · 1