10. In the Cartesian coordinate system $x O y$, it is known that the vertices of the equilateral $\triangle A B C$ (vertices in counterclockwise order) are $A(-2,2)$, vertex $B$ is on the $y$-axis, and $C B \parallel x$-axis; point $D(2,6)$, $E(2,2)$, and the radius of $\odot O$ is 2. Find the radius of the circle that is externally tangent to $\odot O$ and tangent to line segments $A B$ and $D E$ (accurate to $10^{-4}$).
10. From the problem, we know that the equation of the line on which segment $AB$ lies is
$$
y-2=\sqrt{3}(x+2) \text {. }
$$

Therefore, the equation of segment $AB$ is
$$
\sqrt{3} x-y+2 \sqrt{3}+2=0(-2 \leqslant x \leqslant 0) \text {. }
$$

Let the center of the circle that is externally tangent to $\odot O$ and tangent to segments $AB$ and $DE$ be $M\left(x_{0}, y_{0}\right)$, with radius $r$. Thus,
$$
\begin{array}{l}
x_{0}=2-r, M O=2+r . \\
\text { Then }(2+r)^{2}=M O^{2}=x_{0}^{2}+y_{0}^{2}=(2-r)^{2}+y_{0}^{2} \\
\Rightarrow y_{0}=2 \sqrt{2 r} .
\end{array}
$$

Since the distance from the center $M$ to the line $AB$ equals $r$, we have
$$
\begin{aligned}
& \frac{|\sqrt{3}(2-r)-2 \sqrt{2 r}+2 \sqrt{3}+2|}{2}=r \\
& \Rightarrow(2+\sqrt{3}) r+2 \sqrt{2 r}-(4 \sqrt{3}+2)=0 \\
& \Rightarrow \sqrt{r}=\frac{-\sqrt{2}+\sqrt{10 \sqrt{3}+18}}{2+\sqrt{3}} \approx 1.2135 \\
\Rightarrow r & =1.4726 .
\end{aligned}
$$