jmc

Note that the center of the circle is the midpoint of $AB$, call it $M$. When we decrease $x$, the limiting condition is that the circle will eventually be tangent to segment $AD$ at $D$ and segment $BC$ at $C$. That is, $MD\perp AD$ and $MC\perp BC$. From here, we drop the altitude from $D$ to $AM$; call the base $N$. Since $△DNM\sim △ADM$, we have$$\frac{DM}{19/2}=\frac{46}{DM}.$$Thus, $DM=\sqrt{19\cdot 23}$. Furthermore, $x^2=A{M}^2-D{M}^2=46^2-19\cdot 23=\boxed{1679}.$