jmc

The equation can be rewritten as $$\begin{array}{rl}{x}^2-14x+y^2-8y& =-49\\ x^2-14x+49+y^2-8y+16& =16\\ (x-7{)}^2+(y-4{)}^2& =16\end{array}$$Thus, the region is a circle with center $(7,4)$ and radius 4. As $(7,4)$ is on the line $y=x-3$, the line passes through the center of the circle. Hence, half of the area of the circle lies below the line $y=x-3$. The radius of the circle is 4, so the circle has area $16\pi$. Therefore, half the area of the circle is $\boxed{8\pi }$.