9. The polar equation of curve $C$ is $\rho=1+\cos \theta$, and the polar coordinates of point $A$ are $(2,0)$. If curve $C$ rotates once around $A$ in its plane, then the area of the figure swept by it is $\qquad$
Let point $B(\rho, \theta)$ be on the curve $C$, then $|A B|^{2}=\rho^{2}+4-4 \rho \cos \theta$
$$
\begin{array}{l}
=(1+\cos \theta)^{2}-4 \cos \theta(1+\cos \theta) \\
=-3 \cos ^{2} \theta-2 \cos \theta+5=-3\left(\cos \theta+\frac{1}{3}\right)^{2}+\frac{16}{3} .
\end{array}
$$

Thus, when $\cos \theta=-\frac{1}{3}$, $|A B|_{\text {max }}^{2}=\frac{16}{3}$.
Therefore, the area of the figure swept by the curve $C$ is $\frac{16}{3} \pi$.