10. (20 points) Given the parabola $C: y=\frac{1}{2} x^{2}$ and the circle $D: x^{2}+\left(y-\frac{1}{2}\right)^{2}=r^{2}(r>0)$ have no common points, a tangent line is drawn from a point $A$ on the parabola $C$ to the circle $D$, with the points of tangency being $E$ and $F$. As point $A$ moves along the parabola $C$, all points not on the line $E F$ form a planar region $\Omega$. Find the range of the area of region $\Omega$.
10. Substituting the parabola equation into the equation of circle $D$ yields
$$
2 y+\left(y-\frac{1}{2}\right)^{2}=r^{2} \Rightarrow y^{2}+y=r^{2}-\frac{1}{4} \text {. }
$$

From the fact that this equation has no positive real roots, we know
$$
r^{2}-\frac{1}{4}<0 \Rightarrow 0<r<\frac{1}{2} \text {. }
$$

Let $A\left(m, \frac{1}{2} m^{2}\right)$. Then
$$
\begin{array}{l}
l_{E F}: m x+\left(\frac{1}{2} m^{2}-\frac{1}{2}\right)\left(y-\frac{1}{2}\right)=r^{2} \\
\Rightarrow\left(\frac{1}{2} y-\frac{1}{4}\right) m^{2}+x m-\frac{1}{2} y+\frac{1}{4}-r^{2}=0 .
\end{array}
$$

Notice that, regardless of the value of $m$, the line $E F$ does not pass through the point $(x, y)$ if and only if equation (1) has no real roots, i.e.,
$$
\begin{array}{l}
\Delta=x^{2}-4\left(\frac{1}{2} y-\frac{1}{4}\right)\left(-\frac{1}{2} y+\frac{1}{4}-r^{2}\right)<0 \\
\Rightarrow x^{2}+\left(y+r^{2}-\frac{1}{2}\right)^{2}<r^{4} .
\end{array}
$$

This indicates that the locus of point $(x, y)$ is the interior of a circle with center $\left(0, \frac{1}{2}-r^{2}\right)$ and radius $r^{2}$. Therefore, the area of the desired region is $\pi r^{4}$, with the range of values being
$$
\left(0, \frac{\pi}{16}\right) .
$$