jmc

Add $(-6/2{)}^2$ and $(-14/2{)}^2$ to both sides of the equation to get $$(x^2-6x+9)+(y^2-14y+49)=25,$$ which in turn can be rewritten as $(x-3{)}^2+(y-7{)}^2=5^2$. The center of this circle is $(3,7)$, so the line $y=7$ passes through the center of the circle. Hence, the area of the circle that lies below $y=7$ is half the area of the circle. The radius of the circle is $\sqrt{25}=5$, so the circle has area $25\pi$. Therefore, half the area of the circle is $\boxed{\frac{25\pi }{2}}$.