Question

The maximum area of a rectangle inscribed in the circle $(x+1{)}^2+(y-3{)}^2=64$ is

• A. $64$ sq. units
• B. $72$ sq. units
• C. $128$ sq. units
• D. $8$ sq. units

Answer 1

The correct option is C $128$ sq. units

Area of rectangle $ABCD=AB\times BC$
$=2OP\times 2PB$
$=2r\sin \theta \times 2r\cos \theta ,\theta <\frac{\pi }{2}$
$=2r^2\sin 2\theta =128\sin 2\theta =A\left(say\right)$
$\frac{dA}{d\theta }=256\cos 2\theta$
Put $\cos 2\theta =0\Rightarrow \theta =\frac{\pi }{4}$
$\text{Now},\frac{d^{2}A}{d\theta }{|}_{\theta =\pi /4}=-512<0$
$\therefore A$ attains maximum value at $\theta =\frac{\pi }{4}$.
Maximum Area $=128\times \sin (2\cdot \frac{\pi }{4})=128$