jmc

Writing out the equations of the circles, we have that: $$\begin{array}{rl}(x-2{)}^2+(y+1{)}^2& =16\\ (x-2{)}^2+(y-5{)}^2& =10\end{array}$$To solve for the common $y$ value of both $A$ and $B$, we can subtract the two equations to find that $(y+1{)}^2-(y-5{)}^2=6$. Simplifying gives that $(y+1{)}^2-(y-5{)}^2=2y+1+10y-25=12y-24=6,$ so that $y=\frac{30}{12}=\frac{5}{2}$. Substituting back into either of the above circle equations yields that $(x-2{)}^2=\frac{15}{4}$. Thus, $x-2=\pm \frac{\sqrt{15}}{2}$, so $x=2\pm \frac{\sqrt{15}}{2}$. The distance between $A$ and $B$ is simply the difference of their x-coordinates, or $$\left(2+\frac{\sqrt{15}}{2}\right)-\left(2-\frac{\sqrt{15}}{2}\right)=\sqrt{15}.$$Thus $(AB{)}^2=(\sqrt{15}{)}^2=\boxed{15}$.