jmc

Completing the square, the equation of the circle can be rewritten in the form $$(x^2-12x+36)+y^2=64,$$or $(x-6{)}^2+y^2=8^2.$ The center of this circle is $(6,0)$, so both the $x$-axis and the line $y=6-x$ pass through the center of the circle: Since the line $y=6-x$ has slope $-1,$ it makes an angle of $135^{\circ }$ with the positive $x-$axis, so the desired region makes up $\frac{135^{\circ }}{360^{\circ }}=\frac{3}{8}$ of the circle. The radius of the circle is $\sqrt{64}=8$, so the circle has area $8^2\pi =64\pi$. Therefore, the desired area is $64\pi \cdot \frac{3}{8}=\boxed{24\pi }$.