14. In the Cartesian coordinate system $x O y$, point $A(0,3)$, line $L: y=2 x-4$. Let the radius of circle $C$ be 1, and the center $C$ lies on line $L$. If there exists a point $M$ on circle $C$ such that $|M A|=2|M D|$, then the range of the x-coordinate $a$ of the center $C$ is $\qquad$.
Given $B(0,-3), D(0,1), E(0,-1)$, the point $M$ that satisfies $|M A|=2|M D|$ lies on the circle with $B D$ as its diameter, centered at $E$. The necessary and sufficient condition for the existence of point $M$ is $1 \leqslant|E C| \leqslant 3$. As shown in the figure, let $C(a, 2 a-4)$, then $1 \leqslant a^{2}+(2 a-3)^{2} \leqslant 9 \Rightarrow\left\{\begin{array}{l}5 a^{2}-12 a \leqslant 0, \\ 5 a^{2}-12 a+8 \geqslant 0\end{array}\right.$ $\Rightarrow 0 \leqslant a \leqslant \frac{12}{5}$. Therefore, the range of $a$ is $\left[0, \frac{12}{5}\right]$.