Find the area of the region enclosed by the graph of the equation $x^2-14x+3y+70=15+9y-y^2$ that lies below the line $y=x-3$.
First we shift all terms to one side in order to simplify the equation: \begin{align*}
0=x^2-14x+3y+70-15-9y+y^2&\Rightarrow 0=y^2-6y-11x+40
\end{align*}We proceed to complete the square to simplify this expression further. $$0=(y^2-6y+9)-9 + (x^2-\frac{11x}{2})+\frac{11^2}{4}-\frac{11^2}{4}+\frac{40}{4}$$This expression simplifies to $$0=(y-3)^2+(x-\frac{11}{2})^2-\frac{193}{4}=(y-3)^2+(x-\frac{11}{2})^2-\left(\frac{2\sqrt{193}}{4}\right)^2$$Thus we know this equation describes a circular arc with center $\left( \frac{11}{2},3 \right)$ and radius $\frac{2\sqrt{193}}{4}=\frac{\sqrt{193}}{2}$. Now, let's consider the line $y=x-3$. Since we're concerned only with the region below this line, we want to find the arc of the circle below this line. [asy]
draw((4,0)--(12,0),EndArrow);
draw((0,0)--(0,8),EndArrow);
draw((11/2,3)--(11/2,11/2),EndArrow,black+linewidth(1.5));
dot((11/2,3));
fill((11/2+sqrt(193)/2,3)--(11/2+sqrt(193)/2,11/2)--(11/2,sqrt(193)/2+3)--(11/2-sqrt(193)/2,11/2)--(11/2-sqrt(193)/2,3)--(11/2,3)--cycle,gray(.7));
draw(arc((11/2,3),sqrt(193)/2,0,180));
draw((4,0)--(12,12)--(12,0),black+linewidth(1.5));
label("$y=x-3$", (11/2,11/3), NE);
label("$\frac{\sqrt{193}}{2}$", (1.5,1.5), SW);
label("$\left( \frac{11}{2}, 3 \right)$", (11/2, 3), SE);
draw((11/2,3)--(11/2+sqrt(193)/2,3)--(11/2+sqrt(193)/2,sqrt(193)/2+3)--(11/2,sqrt(193)/2+3),red+linewidth(1.5));
label("$\frac{\sqrt{193}}{2}$", (9.2,4.7), ENE);
[/asy] We see that the arc crosses below the line $y=x-3$ at two points. Let $(P,Q)$ be the point where this arc crosses the line and $(11/2+t,3+t)$ be the point which lies on the arc. By the distance formula, we have \begin{align*}
\sqrt{t^2 + (t+3)^2}&=\frac{\sqrt{193}}{2}\\
2t^2+6t+9&= \frac{193}{4}\\
8t^2+24t+36-193&= 0\\
8t^2+24t-157&= 0
\end{align*}Applying the quadratic formula on $8t^2+24t-157=0$, we obtain \begin{align*}
t&=\frac{-24\pm \sqrt{(24)^2-4(8)(-157)}}{16}\\
&=\frac{-24\pm 36}{16}
\end{align*}which implies $t=-9/4,\frac{5}{2}$. For $t=-9/4,$ $(11/2+t,3+t)=(11/2-9/4,3-9/4)=(7/4,3/4)$ which is below the $x-$axis. For $t=5/2,$ $(11/2+t,3+t)=(11/2+5/2,3+5/2)=(16/2,13/2)=(8,13/2)$ which is above the line $y=x-3$. Thus, the region we are concerned with is summarily: [asy]
fill((11/2+sqrt(193)/2,3)--(11/2-9/4,3-9/4)--(4,0)--(12,0)--cycle,gray(.7));
draw((4,0)--(11, 0),black+linewidth(1.5));
draw((0,0)--(0,8),EndArrow);
draw((11/2,3)--(11/2,11/2),EndArrow,black+linewidth(1.5));
draw((11/2,3)--(11/2-9/4,3-9/4)--(4,0),red+linewidth(1.5));
draw((11/2,3)--(11/2+5/2,3+5/2),blue+linetype("5 4"));
fill((4, 0)--(11/2-9/4,3-9/4)--(7/4, 3/4)--cycle, gray(.7));
label("$y=x-3$", (11/2,11/3), NE);
label("$\frac{\sqrt{193}}{2}$", (1.5,1.5), SW);
label("$\left( \frac{11}{2}, 3 \right)$", (11/2, 3), SE);
draw(arc((11/2,3),sqrt(193)/2,0,180));
draw((1/4,3/4)--(-1/4,5/4)--(-9/4, 3-9/4),dotted);
dot((7/4, 3/4));
[/asy]The area of this region is just the area of the region between the circle and the triangle: \begin{align*}
\left(\frac{\pi(\sqrt{193}/2)^2}{2} \right) + \left( \frac{1/4 \cdot 3/4\cdot 7}{3} \right)&= \boxed{\frac{193}{8\pi}+\frac{7}{4}}
\end{align*}
To find the area of the region enclosed by the graph of the equation and the line, we first need to determine the points of intersection between the two curves.
Step 1: Set the two equations equal to each other:
$x^2-14x+3y+70=15+9y-y^2$
$y=x-3$
Step 2: Rearrange the equations to form a quadratic equation in terms of $y$:
$x^2-14x+3y+70+y^2-9y-15=0$
$x^2-14x+3y^2-6y+55=0$
Step 3: Complete the square to simplify the equation:
$x^2-14x+49-49+3y^2-6y+3-3+55=0$
$(x-7)^2+3(y-1)^2+8=0$
Step 4: Since both $(x-7)^2$ and$(y-1)^2$ are non-negative, the equation $(x-7)^2+3(y-1)^2+8=0$ has no real solutions. Hence, the two curves do not intersect.
Thus, the area enclosed by the graph of the equation and the line is 0.