imc

Let $D$ and $E$ be the points of tangency on circles $A$ and $B$ with line $CD$. $AB=8$. Also, let $BC=x$. As $\angle ADC$ and $\angle BEC$ are right angles (a radius is perpendicular to a tangent line at the point of tangency) and both triangles share $\angle ACD$, $△ADC\sim △BEC$. From this we can get a proportion. $\frac{BC}{AC}=\frac{BE}{AD}\to \frac{x}{x+8}=\frac{3}{5}\to 5x=3x+24\to x=\boxed{12}$