problem: What is the area of the region bounded by the graph of the equation $x^2+y^2 + 3 = -4y + 6x-1$?
solution: 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 $\pi r^2 = \pi (3)^2 = \boxed{9\pi}$.