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

Explanation for the correct answer:

Finding the value of $\alpha \beta$

$$\begin{gathered} \alpha =\left|\frac{3\cos \theta +3\sin \theta -2}{\sqrt{2}}\right| \\ \beta =\left|\frac{3\cos \theta +3\sin \theta +2}{\sqrt{2}}\right| \\ \alpha \beta =\left|\frac{{\left(3\cos \theta +3\sin \theta \right)}^2-4}{2}\right|\left[{\left(a+b\right)}^2=a^2+b^2+2ab\right] \\ \alpha \beta =\left|\frac{9+9\sin 2\theta -4}{2}\right|[{\sin }^2\theta +{\cos }^2\theta =1] \\ \alpha \beta =\left|\frac{5+9\sin 2\theta }{2}\right| \\ \alpha \beta \le 7(\because \sin \theta \le 1) \end{gathered}$$

Therefore the correct answer is $\alpha \beta \le 7$