Question

The area common to the circles $r=a\sqrt{2}$ and $r=2a\cos \theta$ is:

• A. $a^2\frac{\pi }{2}$
• B. $a^2\pi$
• C. $a^2(\pi +1)$
• D. $a^2(\pi -1)$

Answer 1

The correct option is D $a^2(\pi -1)$

The equations of the circles are $r=a\sqrt{2}$ ................$\left(1\right)$
and $r=2a\cos \theta$ .................$\left(2\right)$
eqn$\left(1\right)$ represents a circle with centre at $(0,0)$ and radius $a\sqrt{2}$
eqn$\left(2\right)$ represents a circle symmetrical about $OX$ with centre at $(a,0)$ and radius $a$
The circles are shown in fig.At their point of intersection $P$,eliminating $r$ from $\left(1\right)$ and $\left(2\right)$,
$a\sqrt{2}=2a\cos \theta$
$\Rightarrow \cos \theta =\frac{1}{\sqrt{2}}$
or $\theta =\frac{\pi }{4}$
$\therefore$ Required Area$=2\times$area$OAPQ$ (By symmetry)
$=2$(area$OAP+$area$OPQ$)
$=2[\frac{1}{2}{\int }_0^{\frac{\pi }{4}}{r}^{2}d\theta +\frac{1}{2}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{r}^{2}d\theta ]$ for $\left(1\right)$ and $\left(2\right)$ respectively.
$={\int }_0^{\frac{\pi }{4}}{\left(a\sqrt{2}\right)}^{2}d\theta +{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\left(2a\cos \theta \right)}^{2}d\theta$
$=2a^2{\left|\theta \right|}_0^{\frac{\pi }{4}}+4a^2{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\frac{1+\cos \theta }{2}d\theta$
$=2a^2(\frac{\pi }{4}-0)+2a^2{|\theta +\frac{\sin 2\theta }{2}|}_{\frac{\pi }{4}}^{\frac{\pi }{2}}$
$=\frac{\pi a^2}{2}+2a^2(\frac{\pi }{2}-\frac{\pi }{4}-\frac{1}{2})$
$a^2(\pi -1)$