14. In the Cartesian coordinate system $x O y$, point $A(0,3)$, line $l: y=2 x-4$, let the radius of $\odot C$ be 1, and the center of the circle lies on line $l$. If there exists a point $M$ on $\odot C$ such that $|M A|=2|M O|$, then the range of the x-coordinate $a$ of the center $C$ is $\qquad$
14. $\left[0, \frac{12}{5}\right]$.

Notice that, the trajectory equation of point $M$ is
$$
x^{2}+(y+1)^{2}=4 \text {, }
$$

with its center at $B(0,-1)$.
The moving circle center is $C(a, 2 a-4)$, and according to the problem, the distance between the two centers is no greater than the sum of their radii.
$$
\begin{array}{l}
\text { Hence }|B C|=\sqrt{a^{2}+(2 a-4+1)^{2}} \leqslant 3 \\
\Rightarrow 0 \leqslant a \leqslant \frac{12}{5} .
\end{array}
$$

Therefore, the solution is $a \in\left[0, \frac{12}{5}\right]$.