4. The chord $AB=18$ of the sector $OAB$, a circle $C$ with radius 6 is exactly tangent to $OA$, $OB$, and the arc $\widehat{AB}$. Another circle $D$ is tangent to circle $C$, $OA$, and $OB$ (as shown in the figure). Then the radius of circle $D$ is $\qquad$.
4. Solution As shown in the figure, let the radii of $\odot O, \odot D$ be $x, y$ respectively. Then $\alpha C=x-6$.
Since $\triangle O L B \backsim \triangle O M C$.

Then
$$
\frac{O B}{O C}=\frac{L B}{C M},
$$

which means $\frac{x}{x-6}=\frac{9}{6}$.
Solving for $x$ gives $x=18, Q C=12$, $O D=$ $6-y$.
Also, since $\triangle O N D \backsim$
$\triangle O M C$, then
$$
\frac{O D}{O C}=\frac{D N}{C M} \text {, i.e. } \frac{6-y}{12}=\frac{y}{6} .
$$

Solving for $y$ gives $y=2$, i.e., the radius of $\odot D$ is 2.