smc

Label the center of both circles $O$. Label the chord in the larger circle as $\overline{ABCD}$, where $A$ and $D$ are on the larger circle and $B$ and $C$ are on the smaller circle. Construct the radius perpendicular to the chord and label their intersection as $M$. Because a radius that is perpendicular to a chord bisects the chord, $M$ is the midpoint of the chord. Construct segments $\overline{AO}$ and $\overline{BO}$. These are radii with lengths 17 and 19 respectively. Then, use the Pythagorean Theorem. In $△OMA$, we have $$\begin{array}{rl}O{M}^2& =O{A}^2-A{M}^2\\ & =O{A}^2-{\left(\frac{AD}{2}\right)}^2\\ & =19^2-\frac{A{D}^2}{4}.\end{array}$$ In $△OMB$, we have $$\begin{array}{rl}O{M}^2& =O{B}^2-B{M}^2\\ & =O{B}^2-{\left(\frac{BC}{2}\right)}^2\\ & =O{B}^2-{\left(\frac{AD}{4}\right)}^2\\ & =17^2-\frac{A{D}^2}{16}.\end{array}$$ Equating these two expressions, we get $$19^2-\frac{A{D}^2}{4}=17^2-\frac{A{D}^2}{16}$$ and $AD=\boxed{8\sqrt{6}}$.