jmc

We first draw out the figure described in this problem and label the important points with circle $A$ having radius $10$ ft and circle $B$ having radius $17$ feet: Since $\overline{AC}$ is a radius of circle $A$ and $\overline{BC}$ is a radius of circle $B$, we have that $AC=10$ and $BC=17$. Also, since $\overline{DC}$ is a common chord to the two circles, line segment $\overline{AB}$, which connects the centers of the two circles, must both bisect $\overline{DC}$ and be perpendicular to it. We'll call the point of intersection of these two lines point $E$ and since $DC=16$, $\overline{EC}$ must have length $8$.

Now we notice that we have two right triangles $△AEC$ and $△BEC$. Since we know the lengths of $\overline{AC}$ and $\overline{EC}$, we can find the length of $\overline{AE}$ using the Pythagorean Theorem: $$\begin{array}{rl}& A{E}^2+E{C}^2=A{C}^2\\ \Rightarrow & AE=\sqrt{10^2-8^2}=6\end{array}$$ Similarly, we can use the Pythagorean Theorem to find that the length of $\overline{EB}$ is $\sqrt{17^2-8^2}=15$. The length of $\overline{AB}$, the distance between the two centers of the circles, must be the sum of the lengths of $\overline{AE}$ and $\overline{EB}$, which is $6+15=\boxed{21}$ feet.