Question

Find the area of the region enclosed between the two circles $x^2+y^2=4$ and $(x-2)+y^2=4$.

Answer 1

Equation of the given circles are -
$x^2+y^2=4\cdots \left(1\right)$
$(x-2)+y^2=4\cdots \left(2\right)$
on solving the $\left(1\right)$ and $\left(2\right)$ equations, we have
$(x-2{)}^2+y^2=x^2+y^2\Rightarrow x^2-4x+4=x^2$
$\Rightarrow x=1$ which gives $y=\pm \sqrt{3}$

$\therefore$ Required area of the enclosed region $OAC{A}^{\prime }O$ between circles $=2\left(\text{area of the region}ODCAO\right)$
$=2\left(\text{area of the region ODAO}\right)+2\left(\text{area of the region}DCAD\right)$
$={\int }_0^{1}ydx+{\int }_1^{2}ydx=2\left[{\int }_0^1\sqrt{4-(x-2{)}^2}dx+{\int }_1^2\sqrt{4-x^2}dx\right]$
$={\left[\frac{1}{2}\right(x-2)\sqrt{4-(x-2{)}^2}+\frac{1}{2}\times 4{\sin }^{-1}\left(\frac{x-2}{2}\right)]}_0^1+{[x\sqrt{4-x^2}+4{\sin }^{-1}\frac{x}{2}]}_1^2$
$=[(-\sqrt{3}+4{\sin }^{-1}\left(\frac{-1}{2}\right))-4{\sin }^{-1}(-1\left)\right]+[4{\sin }^{-1}1-\sqrt{3}-4{\sin }^{-1}\frac{1}{2}]$
$=[(-\sqrt{3}-4\times \frac{\pi }{6})+4\times \frac{\pi }{2}]+[4\times \frac{\pi }{2}-\sqrt{3}-4\times \frac{\pi }{6}]=(-\sqrt{3}-\frac{2\pi }{3}+2\pi +2\pi -\sqrt{3}-\frac{2\pi }{3})=\frac{8\pi }{3}-2\sqrt{3}$