9. (16 points) Let the moving point $P$ be on the right branch of the hyperbola
$$
\frac{x^{2}}{16}-\frac{y^{2}}{9}=1
$$

(excluding the vertex), $Q$ and $R$ are the left and right foci of the hyperbola, respectively, $S$ is the excenter of $\triangle PQR$ inside $\angle PQR$, and $D(0,1)$. Find the minimum value of $|DS|$.
9. Let the equation of the hyperbola be \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\).
Let \(P\left(x_{0}, y_{0}\right), Q(-c, 0), R(c, 0), S(x, y)\), and \(P R\) intersects \(Q S\) at point \(T\). Let the complex numbers corresponding to points \(P, Q, R, S, T\) be \(p, q, r, s, t\), respectively.
It is easy to see that \(q = -r\).
Let \(|P Q| = u, |P R| = v\). Then
\[
u = e x_{0} + a, \quad v = e x_{0} - a.
\]

By the properties of internal and external angle bisectors, we have
\[
\begin{array}{l}
\frac{|P T|}{|T R|} = \frac{|P Q|}{|Q R|} \Rightarrow \frac{p - t}{t - r} = \frac{u}{2c} \\
\Rightarrow t = \frac{u r + 2c p}{u + 2c}, \\
\frac{|S Q|}{|S T|} = \frac{|P Q|}{|P T|} = \frac{|R Q|}{|R T|} \\
\Rightarrow \frac{|S Q|}{|S T|} = \frac{|P Q| + |R Q|}{|P R|} \\
\Rightarrow \frac{s - q}{s - t} = \frac{u + 2c}{v} \\
\Rightarrow s = \frac{(u + 2c) t - q v}{u + 2c - v}.
\end{array}
\]

Substituting equation (1) into equation (2), we get
\[
\begin{aligned}
s & = \frac{u r + 2c p - q v}{u + 2c - v} = \frac{(u + v) r + 2c p}{2(a + c)} = \frac{e x_{0} r + c p}{a + c} \\
& \Rightarrow (x, y) = \left(\frac{c(e + 1) x_{0}}{a + c}, \frac{c y_{0}}{a + c}\right) = \left(\frac{c x_{0}}{a}, \frac{c y_{0}}{a + c}\right) \\
\Rightarrow & x_{0} = \frac{a x}{c}, \quad y_{0} = \frac{(a + c) y}{c}.
\end{aligned}
\]

Substituting into the hyperbola equation, we get
\[
\frac{x^{2}}{c^{2}} - \frac{(a + c)^{2} y^{2}}{b^{2} c^{2}} = 1.
\]

Thus, the trajectory equation of point \(S\) is
\[
\frac{x^{2}}{25} - \frac{9 y^{2}}{25} = 1.
\]

Let the circle with center \(D\) and radius \(|D S| = r_{0}\) be
\[
x^{2} + (y - 1)^{2} = r_{0}^{2}.
\]

Combining equations (3) and (4), we get
\[
\begin{array}{l}
r_{0}^{2} - (y - 1)^{2} - 9 y^{2} = 25 \\
\Rightarrow 10 y^{2} - 2 y + 26 - r_{0}^{2} = 0.
\end{array}
\]

By the minimality of \(|D S|\), the circle \(\odot D\) is tangent to the hyperbola (3), so
\[
\Delta = 4 - 40(26 - r_{0}^{2}) = 0 \Rightarrow r_{0} = \sqrt{25.9}.
\]