10. Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+y^{2}=1(a>1)$. Given that any circle centered at point $A(0,1)$ intersects the ellipse at most at 3 points, find the range of the eccentricity of the ellipse.
Solve:
As shown in the figure, if the circle with center at point $A(0,1)$ intersects the ellipse at 4 points, by symmetry, there must be two intersection points $P, Q$ on the left side of the $y$-axis, at this time $k_{A P}>k_{A Q}>0$. Let's assume $k_{A P}=k_{1}, k_{A Q}=k_{2}$, and solve the system of equations:
$$
\begin{array}{l}
\left\{\begin{array}{l}
y=k x+1, \\
x^{2}+a^{2} y^{2}=a^{2}
\end{array} \Rightarrow\left(a^{2} k^{2}+1\right) x^{2}+2 a^{2} k x=0\right. \\
\Rightarrow x_{P}=-\frac{2 a^{2} k_{1}}{a^{2} k_{1}^{2}+1}, x_{Q}=-\frac{2 a^{2} k_{2}}{a^{2} k_{2}^{2}+1} .
\end{array}
$$

Thus, $|A P|=|A Q| \Rightarrow \sqrt{k_{1}^{2}+1} \cdot \frac{2 a^{2} k_{1}}{a^{2} k_{1}^{2}+1}=\sqrt{k_{2}^{2}+1} \cdot \frac{2 a^{2} k_{2}}{a^{2} k_{2}^{2}+1}$
$$
\begin{array}{l}
\Rightarrow \frac{\sqrt{k_{1}^{2}+1} \cdot k_{1}}{a^{2} k_{1}^{2}+1}=\frac{\sqrt{k_{2}^{2}+1} \cdot k_{2}}{a^{2} k_{2}^{2}+1} \Rightarrow \frac{\sqrt{1+\frac{1}{k_{1}^{2}}}}{a^{2}+\frac{1}{k_{1}^{2}}}=\frac{\sqrt{1+\frac{1}{k_{2}^{2}}}}{a^{2}+\frac{1}{k_{2}^{2}}} \\
\Rightarrow\left(a^{2}+\frac{1}{k_{1}^{2}}\right)^{2}\left(1+\frac{1}{k_{2}^{2}}\right)=\left(a^{2}+\frac{1}{k_{2}^{2}}\right)^{2}\left(1+\frac{1}{k_{1}^{2}}\right) \\
\Rightarrow\left(a^{4}+\frac{2 a^{2}}{k_{1}^{2}}+\frac{1}{k_{1}^{4}}\right)\left(1+\frac{1}{k_{2}^{2}}\right)=\left(a^{4}+\frac{2 a^{2}}{k_{2}^{2}}+\frac{1}{k_{2}^{4}}\right)\left(1+\frac{1}{k_{1}^{2}}\right) \\
\Rightarrow a^{4}+\frac{2 a^{2}}{k_{1}^{2}}+\frac{1}{k_{1}^{4}}+\frac{a^{4}}{k_{2}^{2}}+\frac{2 a^{2}}{k_{1}^{2} k_{2}^{2}}+\frac{1}{k_{1}^{4} k_{2}^{2}}=a^{4}+\frac{2 a^{2}}{k_{2}^{2}}+\frac{1}{k_{2}^{4}}+\frac{a^{4}}{k_{1}^{2}}+\frac{2 a^{2}}{k_{1}^{2} k_{2}^{2}}+\frac{1}{k_{1}^{2} k_{2}^{4}} \\
\Rightarrow \frac{2 a^{2}}{k_{1}^{2}}-\frac{2 a^{2}}{k_{2}^{2}}+\frac{a^{4}}{k_{2}^{2}}-\frac{a^{4}}{k_{1}^{2}}=\frac{1}{k_{2}^{4}}-\frac{1}{k_{1}^{4}}+\frac{1}{k_{1}^{2} k_{2}^{4}}-\frac{1}{k_{1}^{4} k_{2}^{2}} \\
\Rightarrow\left(2 a^{2}-a^{4}\right)\left(\frac{1}{k_{1}^{2}}-\frac{1}{k_{2}^{2}}\right)=-\left(\frac{1}{k_{1}^{2}}+\frac{1}{k_{2}^{2}}+\frac{1}{k_{1}^{2} k_{2}^{2}}\right)\left(\frac{1}{k_{1}^{2}}-\frac{1}{k_{2}^{2}}\right)
\end{array}
$$
$\Rightarrow 2 a^{2}-a^{4}=-\left(\frac{1}{k_{1}^{2}}+\frac{1}{k_{2}^{2}}+\frac{1}{k_{1}^{2} k_{2}^{2}}\right)2 \Rightarrow a>\sqrt{2}$. The above process is also valid in reverse, so the circle with center at point $A(0,1)$ intersects the ellipse at most at 3 points $\Leftrightarrow 1<a \leqslant \sqrt{2}$.
Therefore, the eccentricity of the ellipse $e=\sqrt{1-\frac{1}{a^{2}}} \in\left(0, \frac{\sqrt{2}}{2}\right]$.