jmc

We rewrite the equation as $x^2-6x+y^2+4y=-4$ and then complete the square, resulting in $(x-3{)}^2-9+(y+2{)}^2-4=-4$, or $(x-3{)}^2+(y+2{)}^2=9$. This is the equation of a circle with center $(3,-2)$ and radius 3, so the area of this region is $\pi r^2=\pi (3{)}^2=\boxed{9\pi }$.