The area of the region enclosed between the two circles $x^2+y^2 = 1 $ and $(x-1)^2 +y^2 =1$ is

$\left(\frac{2\pi }{3} -\frac{\sqrt{3}}{2}\right)$ units

The correct answer is Option (3) → $\left(\frac{2\pi }{3} -\frac{\sqrt{3}}{2}\right)$ sq. units

By symmetry I = II = III = IV

finding intersection points

$x^2+y^2=1$

$(x-1)^2+y^2=1$

$⇒x^2=(x-1)^2$

so $x=\frac{1}{2}$

so area required = $4 ×area\,I$

$=4 ×\int\limits_{1/2}^1\sqrt{1-x^2}dx$

$=4\left[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}\sin^{-1}x\right]_{1/2}^1$

$=\frac{2\pi }{3} -\frac{\sqrt{3}}{2}$ sq. units