9. (14 points) As shown in Figure 1, given
$\odot P: x^{2}+y^{2}=2 x$, parabola $S: y^{2}=4 x$, a line $l$ is drawn through the center $P$ of the circle, intersecting the two curves at four points, which are sequentially labeled as $A, B, C, D$ from top to bottom. If the lengths of segments $A B, B C, C D$ form an arithmetic sequence in this order, find the equation of line $l$.
Given the conditions:
\[
\odot P:(x-1)^{2}+y^{2}=1,
\]
the diameter length \( |B C|=2 \), and the center \( P(1,0) \).
Let the line \( l: k y=x-1 \), i.e., \( x=k y+1 \).
Substituting into the parabola equation gives \( y^{2}=4 k y+4 \).
Let \( A\left(x_{1}, y_{1}\right), D\left(x_{2}, y_{2}\right) \). Then
\[
\begin{array}{l}
\left\{\begin{array}{l}
y_{1}+y_{2}=4 k, \\
y_{1} y_{2}=-4 \\
\Rightarrow\left(y_{1}-y_{2}\right)^{2}=\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}
\end{array}\right. \\
\Rightarrow|A D|^{2}=\left(y_{1}-y_{2}\right)^{2}\left(1+k^{2}\right)=16\left(k^{2}+1\right)^{2} \\
\Rightarrow A D=4\left(k^{2}+1\right) . \\
\text { Also, } 2|B C|=|A B|+|C D|=|A D|-|B C| \\
\Rightarrow|A D|=3|B C|=6 \\
\Rightarrow 4\left(k^{2}+1\right)=6 \\
\Rightarrow k= \pm \frac{\sqrt{2}}{2} .
\end{array}
\]

Therefore, the line \( l: x=\frac{\sqrt{2}}{2} y+1 \) or \( x=-\frac{\sqrt{2}}{2} y+1 \).