10. (20 points) As shown in Figure 1, the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>$ $b>0)$ has an inscribed circle $x^{2}+y^{2}=b^{2}$. A tangent line to the circle intersects the ellipse at points $A$ and $B$, and the tangent point $Q$ of line $AB$ is on the left side of the $y$-axis. $F$ is the right focus of the ellipse. Find the perimeter of $\triangle ABF$.
10. Note that,
$$
\begin{array}{l}
|A F|=\sqrt{(x-c)^{2}+y^{2}} \\
=\sqrt{x^{2}-2 c x+c^{2}+b^{2}-\frac{b^{2}}{a^{2}} \cdot x^{2}} \\
=\sqrt{\frac{c^{2}}{a^{2}} \cdot x^{2}-2 c x+a^{2}}=a-\frac{c}{a} \cdot x(\text { because } x<a), \\
|A Q|=\sqrt{A O^{2}-O Q^{2}}=\sqrt{x^{2}+y^{2}-b^{2}} \\
=\sqrt{x^{2}+b^{2}-\frac{b^{2}}{a^{2}} \cdot x^{2}-b^{2}} \\
=\sqrt{\frac{c^{2}}{a^{2}} \cdot x^{2}}=\frac{c}{a} \cdot x .
\end{array}
$$

Thus, $|A Q|+|A F|=\frac{c}{a} \cdot x+a-\frac{c}{a} \cdot x=a$.
Similarly, $|B Q|+|B F|=a$.
Therefore, the perimeter of $\triangle A B F$ is $2 a$.