imc

Let $BX=q$, $CX=p$, and $AC$ meets the circle at $Y$ and $Z$, with $Y$ on $AC$. Then $AZ=AY=86$. Using the Power of a Point (Secant-Secant Power Theorem), we get that $p(p+q)=11(183)=11\ast 3\ast 61$. We know that $p+q>p$, so $p$ is either $3$, $11$, or $33$. We also know that $p>11$ by the triangle inequality on $△ACX$. Thus, $p$ is $33$ so we get that $BC=p+q=\boxed{61}$.