The area bounded by curves $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$ is

Given circles are $x^{2}+y^{2}=1 \ldots$ (i) and $(x-1)^{2}+y^{2}=1 \ldots$ (ii) Centre of (i) is $\mathrm{O}(0,0)$ and radius $=1$
Both these circle are symmetrical about $x$-axis solving (i) and (ii), we get, $-2 x+1=0$ $\Rightarrow x=\frac{1}{2}$ $\begin{array}{l} \text { then } y^{2}=1-\left(\frac{1}{2}\right)^{2}=34 \\ \Rightarrow y=\frac{\sqrt{3}}{2} \end{array}$ $\therefore$ The points of intersection are $\mathrm{P}\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ and $\mathrm{Q}\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$ It is clear from the figure that the shaded portion in region whose area is required. $\therefore$ Required area = area OQAPO $=2 \times$ area of the region OLAP $=2 \times($ area of the region $O L P O+$ area of $L A P L$ ) $=2\left[\int_{0}^{\frac{1}{2}} \sqrt{1-(x-1)^{2}} d x+\int_{\frac{1}{2}}^{1} \sqrt{1-x^{2}} \mathrm{dx}\right]$ $=2\left[\frac{(x-1) \sqrt{1-(x-1)^{2}}}{2}+\frac{1}{2} \sin ^{-1}(x-1)\right]_{0}^{\frac{1}{2}}+2\left[\frac{x \sqrt{1-x^{2}}}{2}+\frac{1}{2} \sin ^{-1} x\right]_{\frac{1}{2}}^{1}$ $=-\frac{1}{2} \cdot \frac{\sqrt{3}}{2}+\sin ^{-1}\left(\frac{-1}{2}\right)-\sin ^{-1}(-1)+0+\sin ^{-1}(1)-\left(\frac{1}{2} \cdot \frac{\sqrt{3}}{2}+\sin ^{-1}\left(\frac{1}{2}\right)\right)$ $=\left(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\right)$ sq. units.