jmc

Call the center of the circle $O$. By drawing the lines from $O$ tangent to the sides and from $O$ to the vertices of the quadrilateral, four pairs of congruent right triangles are formed. Thus, $\angle AOP+\angle POB+\angle COQ+\angle QOD=180$, or $(\arctan (\frac{19}{r})+\arctan (\frac{26}{r}))+(\arctan (\frac{37}{r})+\arctan (\frac{23}{r}))=180$. Take the $\tan$ of both sides and use the identity for $\tan (A+B)$ to get$$\tan (\arctan (\frac{19}{r})+\arctan (\frac{26}{r}))+\tan (\arctan (\frac{37}{r})+\arctan (\frac{23}{r}))=n\cdot 0=0.$$ Use the identity for $\tan (A+B)$ again to get$$\frac{\frac{45}{r}}{1-19\cdot \frac{26}{r^2}}+\frac{\frac{60}{r}}{1-37\cdot \frac{23}{r^2}}=0.$$ Solving gives $r^2=\boxed{647}$.