jmc

Let $\overline{BD}$ be an altitude of the isosceles $△ABC$, and let $O$ denote the center of the circle with radius $r$ that passes through $A$, $B$, and $C$, as shown.



Then $$BD=\sqrt{3^2-1^2}=2\sqrt{2}\text{and}OD=2\sqrt{2}-r.$$ Since $△ADO$ is a right triangle, we have $$r^2=1^2+{\left(2\sqrt{2}-r\right)}^2=1+8-4\sqrt{2}r+r^2,\text{and}r=\frac{9}{4\sqrt{2}}=\frac{9}{8}\sqrt{2}.$$ As a consequence, the circle has area $${\left(\frac{9}{8}\sqrt{2}\right)}^2\pi =\boxed{\frac{81}{32}\pi }.$$