smc

Let the circle have center $O$, let the point of tangency be point $T$, and let $B$ be the intersection of $\overline{OA}$ with the circle, as in the diagram. By the definitions of a circle and a tangent to a circle, we know that $OB=OT=r$ and $\overline{OT}\perp \overline{TA}$. By the Pythagorean Theorem, $OA=\sqrt{\frac{25r^2}{9}}=\frac{5}{3}r$. Because the shortest segment from an external point to a circle lies on the line connecting that point to the center of the circle, our desired distance is $AB$. Because $OA=\frac{5}{3}r$ and $OB=r$, $AB=\frac{5}{3}r-r=\frac{2}{3}r=\frac{(4/3)r}{2}=\boxed{\frac{1}{2}l}$.