smc

Draw radii $OE$ and $OD$ to the axes, and label the point of tangency to triangle $ABC$ point $F$. Let the radius of the circle $O$ be $r$. Square $OEAD$ has side length $r$. Because $BD$ and $BF$ are tangents from a common point $B$, $BD=BF$. $AD=AB+BD$ $r=1+BD$ $r=1+BF$ Similarly, $CF=CE$, and we can write: $AE=AC+CE$ $r=\sqrt{3}+CF$ Equating the radii lengths, we have $1+BF=\sqrt{3}+CF$ This means $BF-CF=\sqrt{3}-1$ $BF+CF=2$ by the 30-60-90 triangle. Therefore, $2BF=2+\sqrt{3}-1$, and we get $BF=\frac{1}{2}+\frac{\sqrt{3}}{2}$ The radius of the circle is $AD$, which is $BF+1=\frac{3}{2}+\frac{\sqrt{3}}{2}$ Using decimal approximations, $r\approx 1.5+\frac{1.73^{+}}{2}\approx 2.37$, and the answer is $\boxed{2.37}$.