Question

AB is the chord of contact of tangents drawn from a point $(6,8)$ to the circle $x^2+y^2=r^2.$ If the area of the triangle $PAB$ be maximum, then radius $r$ of the circle is ?

• A. $5$
• B. $5\sqrt{3}$
• C. $10$
• D. $8$

Answer 1

The correct option is A $5$

Let $\angle APB=2\theta ,CP=\sqrt{36+64}=10$
$\therefore r=10\sin \theta$ and $AP=10\cos \theta$
$PM=CP-CM=10-r\sin \theta$ A=Area of $\Delta APB=2\Delta APM$
$A=2.\frac{1}{2}AM.PM=r\cos \theta (10-r\sin \theta )$
Now put $r=10\sin \theta$
$A=100\left[\sin \theta \cos \theta (1-{\sin }^2\theta )\right]=100\left[\sin \theta {\cos }^3\theta \right]$ ...(1)
Now A will be maximum when $z=\sin \theta {\sin }^3\theta$ is max.
$\therefore \frac{dz}{d\theta }={\cos }^4\theta -3{\sin }^2\theta {\cos }^2\theta =0$
or ${\cos }^2({\cos }^2\theta -3{\sin }^2\theta )=0$
$\therefore \cos \theta =0$ or ${\tan }^2\theta =\frac{1}{3}\therefore \theta ={90}^{\circ }$ or $\theta ={30}^{\circ }$
At $\theta ={30}^{\circ }$
$\frac{d^{2}z}{d{\theta }^2}=-2\cos \theta \sin \theta ({\cos }^2\theta -3{\sin }^2\theta )+{\cos }^2\theta [-2\cos \theta \sin \theta -6\sin \theta \cos \theta ]$
$=0-8\sin \theta {\cos }^3\theta =-ive$ for $\theta ={30}^{\circ }$
$\therefore$ z is max. at $\theta ={30}^0$ and hence, area A is maximum.
$\therefore$ Hence, $r=10\sin \theta =10\sin {30}^{\circ }=5$