smc

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 $(CX,CX+XB)$: $$(1,2013)(3,671)(11,183)(33,61).$$ However, by the Triangle Inequality on $△ACX,$ we see that $CX>13.$ This implies that we must have $CX+XB=\boxed{61}.$