Question

From a fixed point $P$ on the circumference of a circle of radius $a,$ the perpendicular $PR$ is drawn to the tangent at $Q$ (a variable point), then the maximum area to the $△PQR$ is

• A. $3\sqrt{3}{a}^2$
• B. $3\sqrt{3}{a}^2/2$
• C. $3\sqrt{3}{a}^2/4$
• D. $3\sqrt{3}{a}^2/8$

Answer 1

Answer: (B)

$3\sqrt{3}{a}^2/2$

Let the circle be $x^2+y^2=a^2.$ We take point P at (a,0) and Q at $(a\cos \theta ,a\sin \theta ).$
Equation of tangent at Q is easily seen to be $x\cos \theta +y\sin \theta =a....\left(1\right)$
Let PR be perpendicular from P on (1).
Then $PR=\frac{a-a\cos \theta }{\sqrt{{\cos }^2\theta +{\sin }^2\theta }}=2a{\sin }^2\frac{\theta }{2}$
Also $PQ=\sqrt{{(a-a\cos \theta )}^2+a^2{\sin }^2\theta }$ $=\sqrt{a^2(1+{\cos }^2\theta -2\cos \theta +{\sin }^2\theta )}=a\sqrt{2(1-\cos \theta )}=2a\sin \frac{\theta }{2}.$
Hence $QR=\sqrt{P{Q}^2-P{R}^2}$ $=\sqrt{4a^2{\sin }^2\left(\theta /2\right)-4a^2{\sin }^4\left(\theta /2\right)}=2a\sin \left(\theta /2\right)\cos \left(\theta /2\right).$
If A denotes the area of $\Delta PQR,$ then

$A=\frac{1}{2}PR.QR=\frac{1}{2}.2a{\sin }^2\left(\frac{\theta }{2}\right).2a\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)=\frac{1}{2}{a}^2(1-\cos \theta )\sin \theta =\frac{1}{2}{a}^2(\sin \theta -\frac{1}{2}\sin 2\theta )$

For maxima or minima, we have

$\frac{dA}{d\theta }=\frac{1}{2}{a}^2(\cos \theta -\cos 2\theta )=0.$
This gives $\cos 2\theta =\cos (2\pi -2\theta )$ or $\theta =2\pi -2\theta$ i.e. $\theta =\frac{2\pi }{3}$

$\frac{d^{2}A}{d{\theta }^2}=\frac{1}{2}{a}^2(-\sin \theta +2\sin 2\theta )=\frac{1}{2}{a}^2(\frac{-\sqrt{2}}{2}-\frac{2\sqrt{3}}{2})<0$ at $\theta =\frac{2\pi }{3}.$
Hence A is maximum when $\theta =\frac{2\pi }{3}.$
$\therefore$ Maximum value of $A=\frac{1}{2}{a}^2[\sin \frac{2\pi }{3}-\frac{1}{2}\sin \frac{4\pi }{3}]$ $=\frac{1}{2}{a}^2[\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{4}]=\frac{3\sqrt{3}}{8}{a}^2$
Ans: B