In $\bigtriangleup ABC$, $AB = 86$, and $AC = 97$. A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$. Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$?

$\textbf{(A)} \ 11 \qquad  \textbf{(B)} \ 28 \qquad  \textbf{(C)} \ 33 \qquad  \textbf{(D)} \ 61 \qquad  \textbf{(E)} \ 72$
Solution 1 (Diophantine PoP)

Let circle $A$ intersect $AC$ at $D$ and $E$ as shown. We apply Power of a Point on point $C$ with respect to circle $A.$ This yields the diophantine equation 
\[CX \cdot CB = CD \cdot CE\]
\[CX(CX+XB) = (97-86)(97+86)\]
\[CX(CX+XB) = 3 \cdot 11 \cdot 61.\]
Since lengths cannot be negative, we must have $CX+XB \ge CX.$ This generates the four solution pairs for $(CX,CX+XB)$: \[(1,2013) \qquad (3,671) \qquad (11,183) \qquad (33,61).\]
However, by the Triangle Inequality on $\triangle ACX,$ we see that $CX>13.$ This implies that we must have $CX+XB= \boxed{\textbf{(D) }61}.$
(Solution by unknown, latex/asy modified majorly by samrocksnature)

Solution 2
Let $BX = q$, $CX = p$, and $AC$ meet the circle at $Y$ and $Z$, with $Y$ on $AC$.  Then $AZ = AY = 86$.  Using the Power of a Point, we get that $p(p+q) = 11(183) = 11 * 3 * 61$.  We know that $p+q>p$, and that $p>13$ by the triangle inequality on $\triangle ACX$.  Thus, we get that $BC = p+q = \boxed{\textbf{(D) }61}$

Solution 3
Let $x$ represent $CX$, and let $y$ represent $BX$. Since the circle goes through $B$ and $X$, $AB = AX = 86$.
Then by Stewart's Theorem,
$xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.$
$x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x$
$x^2 + xy + 86^2 = 97^2$
(Since $y$ cannot be equal to $0$, dividing both sides of the equation by $y$ is allowed.)
$x(x+y) = (97+86)(97-86)$
$x(x+y) = 2013$
The prime factors of $2013$ are $3$, $11$, and $61$. Obviously, $x < x+y$. In addition, by the Triangle Inequality, $BC < AB + AC$, so $x+y < 183$. Therefore, $x$ must equal $33$, and $x+y$ must equal $\boxed{\textbf{(D) }61}$