6. Given circle $\Gamma:(x-1)^{2}+y^{2}=r^{2}(r>0)$, parabola $G: y^{2}=4 x$. A line $l$ passing through the point $(1,0)$ intersects the circle $\Gamma$ at points $C$ and $D$ (point $C$ is above the $x$-axis), and intersects the parabola $G$ at points $A$ and $B$ (point $A$ is above the $x$-axis). If there are three lines $l$ such that $|A C|=$ $|B D|$, then the range of $r$ is $\qquad$
6. $(2,+\infty)$.
(1) When the slope of line $l$ does not exist, $|A C|=$ $|B D|$ holds.
(2) When the slope of line $l$ exists, let
$$
\begin{array}{l}
l: y=k(x-1)(k \neq 0) . \\
\text { Let } A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right) . \\
\text { From }\left\{\begin{array}{l}
y=k(x-1), \\
y^{2}=4 x
\end{array}\right. \\
\Rightarrow k^{2} x^{2}-\left(2 k^{2}+4\right) x+k^{2}=0 \\
\Rightarrow x_{1}+x_{2}=\frac{2 k^{2}+4}{k^{2}} . \\
\text { Also }|A C|=\left|x_{1}+1-r\right|, \\
|B D|=\left|r-\left(x_{2}+1\right)\right| .
\end{array}
$$

From $|A C|=|B D|$, we get
$$
x_{1}+x_{2}=2 r-2 \text {. }
$$

From equations (1) and (2), we have $k^{2}=\frac{2}{r-2}>0$.
Thus, $r>2$.