smc

Let the center of the smaller circle be $A$ and the center of the larger circle be $B$. Draw $\overline{AC}$ and $\overline{BD}$, as in the diagram. We know that $\overline{AC}$ and $\overline{BD}$ are perpendicular to $\overline{CD}$ because $\overline{CD}$ is tangent to both circles. From the problem, we know $CD=24$. Draw $\overline{AE}\perp \overline{BD}$ with $E$ on $\overline{BD}.$ $ABDC$ is a rectangle, so, knowing the radii of the circles, we see that $BE=BD-ED=BD-AC=14-4=10.$ Points $A,E,$ and $B$ form a right triangle with legs $10$ and $24.$ We are looking for $AB,$ which is $\sqrt{10^2+24^2}=26$ by the Pythagorean Theorem. Thus, our answer is $\boxed{26}$.