Consider the curves, $x^2+y^2 =1$ and $(x-1)^2 +y^2=1.$ Area enclosed by the two curves is :

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

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

by symmetry

area I = area II = area III = area IV

finding intersection 

$x^2+y^2=1$

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

$⇒x^2=(x-1)^2$

$x-1=x$

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

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

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

$=π-\frac{\sqrt{3}}{2}-\frac{π}{3}=\frac{2π}{3}-\frac{\sqrt{3}}{2}$