jmc

Above is the diagram implied by the problem, with some added lines ($O$ is the center of the circle). Since $△AMB$ is a 30-60-90 triangle and $AB=9$, $AM=4.5\sqrt{3}$. Since $AO$ is $2/3$ of $AM$, $AO=3\sqrt{3}$. Thus, the area of the circle is $\pi (3\sqrt{3}{)}^2=\boxed{27\pi }$.