imc

Let $△ABC$ have vertex $A$ and center $O$, with foot of altitude from $A$ intersecting $BC$ at $D$. Then by the Pythagorean Theorem (with radius $r$ and height $OD=h$) on $△OBD$ and $△ABD$ $$\begin{array}{rl}{h}^2+1& =r^2\\ (h+r{)}^2+1& =9\end{array}$$ Substituting and solving gives $r=\frac{9}{4\sqrt{2}}$. Then the area of the circle is $r^2\pi ={\left(\frac{9}{4\sqrt{2}}\right)}^2\pi =\boxed{\frac{81}{32}\pi }$.