Solution:
The area bounded by the curves $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$ is reprsented by the shaded area as

On solving the equations, $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$, we get the point of
intersection as $A\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ and $\left.B^{\left(\frac{1}{2}\right.},-\frac{\sqrt{3}}{2}\right)$.
It is observed that the required area is symmetrical about $x$-axis.
$\therefore$ Area $\mathrm{OBCAO}=2 \times$ Area $\mathrm{OCAO}$
Join $A B$, which intersects $O C$ at $M$, such that AM is perpendicular to $O C$.
Coordinates of $\mathrm{M}$ are $\left(\frac{1}{2}, 0\right)$.
$\begin{aligned}
\Rightarrow \operatorname{Area} O C A O &=\text { Area } \mathrm{OMAO}+\text { Area MCAM } \\
&=\left[\int_{0}^{\frac{1}{2}} \sqrt{1-(x-1)^{2}} d x+\int_{\frac{1}{2}}^{1} \sqrt{1-x^{2}} d x\right] \\
&=\left[\frac{x-1}{2} \sqrt{1-(x-1)^{2}}+\frac{1}{2} \sin ^{-1}(x-1)\right]_{0}^{\frac{1}{2}}+\left[\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x\right]_{\frac{1}{2}}^{1} \\
&=\left[-\frac{1}{4} \sqrt{1-\left(-\frac{1}{2}\right)^{2}}+\frac{1}{2} \sin ^{-1}\left(\frac{1}{2}-1\right)-\frac{1}{2} \sin ^{-1}(-1)\right]+\\
&=\left[-\frac{\sqrt{3}}{8}+\frac{1}{2}\left(-\frac{\pi}{6}\right)-\frac{1}{2}\left(-\frac{\pi}{2}\right)\right]+\left[\frac{1}{2}\left(\frac{\pi}{2}\right)-\frac{\sqrt{3}}{8}-\frac{1}{2}\left(\frac{\pi}{6}\right)\right] \\
&=\left[-\frac{\sqrt{3}}{4}-\frac{\pi}{12}+\frac{\pi}{4}+\frac{\pi}{4}-\frac{\pi}{12}\right] \\
&=\left[-\frac{\sqrt{3}}{4}-\frac{\pi}{6}+\frac{\pi}{2}\right] \\
&=\left[\frac{2 \pi}{6}-\frac{\sqrt{3}}{4}\right]
\end{aligned}$
$2 \times\left(\frac{2 \pi}{6}-\frac{\sqrt{3}}{4}\right)=\left(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\right)_{\text {units }}$