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 $\boxed{D}$.
A circle with center $O$ is tangent to the coordinate axes and to the hypotenuse of the $30^\circ$-$60^\circ$-$90^\circ$ triangle $ABC$ as shown, where $AB=1$. To the nearest hundredth, what is the radius of the circle?


$\textbf{(A)}\ 2.18\qquad\textbf{(B)}\ 2.24\qquad\textbf{(C)}\ 2.31\qquad\textbf{(D)}\ 2.37\qquad\textbf{(E)}\ 2.41$