smc

Note as above that ABCD must be tangential to obtain the circle with maximal radius. Let $E$, $F$, $G$, and $H$ be the points on $AB$, $BC$, $CD$, and $DA$ respectively where the circle is tangent. Let $\theta =\angle BAD$ and $\alpha =\angle ADC$. Since the quadrilateral is cyclic(because we want to maximize the circle, so we set the quadrilateral to be cyclic), $\angle ABC=180^{\circ }-\alpha$ and $\angle BCD=180^{\circ }-\theta$. Let the circle have center $O$ and radius $r$. Note that $OHD$, $OGC$, $OFB$, and $OEA$ are right angles. Hence $FOG=\theta$, $GOH=180^{\circ }-\alpha$, $EOH=180^{\circ }-\theta$, and $FOE=\alpha$. Therefore, $AEOH\sim OFCG$ and $EBFO\sim HOGD$. Let $x=CG$. Then $CF=x$, $BF=BE=9-x$, $GD=DH=7-x$, and $AH=AE=x+5$. Using $AEOH\sim OFCG$ and $EBFO\sim HOGD$ we have $r/(x+5)=x/r$, and $(9-x)/r=r/(7-x)$. By equating the value of $r^2$ from each, $x(x+5)=(7-x)(9-x)$. Solving we obtain $x=3$ so that $r=\boxed{2\sqrt{6}}$.