4. Let the line $l$ passing through the point $M(2,0)$ intersect the parabola $y^{2}=4 x$ at points $A$ and $B$, and intersect the circle $(x-4.5)^{2}+y^{2}=16$ at points $C$ and $D$. If $|A C|=|B D|$ and $|A B| \neq|C D|$, then the equation of the line $l$ is . $\qquad$
4. $x=2, y-2 x+4=0, y+2 x-4=0$.

Among the 24 permutations of $A, B, C, D$, the 8 that satisfy $|A C|=|B D|$ and $|A B| \neq|C D|$ are as follows:
$A, C, D, B ; \quad A, D, C, B ; \quad B, C, D, A$;
$B, D, C, A ; \quad C, A, B, D ; \quad D, A, B, C$;
$C, B, A, D ; \quad D, B, A, C$.
Their common feature is that the midpoints of segments $A B$ and $C D$ coincide.
Let the line $l: x=m y+2$.
By combining the line $l$ with the parabola $y^{2}=4 x$ and eliminating $x$, we obtain a quadratic equation in $y$:
$$
y^{2}-4 m y-8=0 \text {. }
$$

Thus, the y-coordinate of the midpoint of $A B$ is $y_{0}=2 m$.
Similarly, the y-coordinate of the midpoint of $C D$ is
$$
y_{0}^{\prime}=\frac{5 m}{2\left(m^{2}+1\right)} \text {. }
$$

Since the midpoints of $A B$ and $C D$ coincide, we have
$$
2 m=\frac{5 m}{2\left(m^{2}+1\right)} \text {. }
$$

Thus, $m\left(4 m^{2}-1\right)=0 \Rightarrow m=0, \pm \frac{1}{2}$.
Therefore, there are three lines $l$ that satisfy the conditions:
$$
x=2, y-2 x+4=0, y+2 x-4=0 \text {. }
$$