9. (16 points) Draw any line $l$ through the point $(0, b)$, intersecting the parabola $y=x^{2}$ at points $A$ and $B$. If there always exists another point $C$ on the parabola, different from $A$ and $B$, such that $C A \perp C B$, find the range of the real number $b$
9. Obviously, the line $l$ must have a slope, let its slope be $k$.
Then the line $l: y=k x+b$.
Let point $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
Solving the system $\left\{\begin{array}{l}y=x^{2}, \\ y=k x+b\end{array}\right.$
$$
\Rightarrow x^{2}-k x-b=0 \text {. }
$$

By the given conditions, we know that equation (1) has two distinct roots $x_{1} 、 x_{2}$.
Let point $C\left(t, t^{2}\right)$. Clearly, $t \neq x_{1}$.
Thus, $k_{A C}=\frac{y_{1}-t^{2}}{x_{1}-t}=x_{1}+t$.
Similarly, $k_{B C}=x_{2}+t$.
By $C A \perp C B \Rightarrow k_{A C} k_{B C}=-1$
$\Rightarrow\left(t+x_{1}\right)\left(t+x_{2}\right)=-1$.
Substituting into equation (1), using the relationship between roots and coefficients, we get
$$
t^{2}+k t-b+1=0 \text {. }
$$

Therefore, under the condition that the line and the parabola have two distinct intersection points (the discriminant of equation (1) $\Delta_{1}=k^{2}+4 b>0$), equation (2) must have a solution $\left(\Delta_{2}=k^{2}+4(b-1) \geqslant 0\right)$, which can only be $b \geqslant 1$.
Conversely, when $b \geqslant 1$, equation (2) must have a solution.

In summary, $b \in[1,+\infty)$.