4. In the Cartesian coordinate plane $x O y$,
$\odot O: x^{2}+y^{2}=1$,
$\odot O_{1}:(x-3)^{2}+y^{2}=4$.
A tangent line is drawn from a point $M$ on the left half-axis of the $x$-axis to $\odot O$, touching $\odot O$ at point $A$, and intersecting $\odot O_{1}$ at points $B$ and $C$. If $A B=B C$, then the coordinates of point $M$ are $\qquad$
4. $(-4,0)$.

As shown in Figure 2, draw $O_{1} H \perp B C$ at point $H$, then $H$ is the midpoint of $B C$.
Let $|O M|=a$, then $M A=\sqrt{a^{2}-1}$.
Thus, $\left|O_{1} H\right|=\frac{a+3}{a}, \frac{|A H|}{3}=\frac{\sqrt{a^{2}-1}}{a}$
$\Rightarrow|A H|=\frac{3 \sqrt{a^{2}-1}}{a}$,
$$
|B H|=\sqrt{2^{2}-\left|O_{1} H\right|^{2}}=\sqrt{4-\left(\frac{a+3}{a}\right)^{2}} .
$$

From the given conditions and $|B H|=\frac{1}{2}|B C|$ we know
$$
\begin{array}{l}
|B H|=\frac{1}{3}|A H| \Rightarrow \sqrt{4-\left(\frac{a+3}{a}\right)^{2}}=\frac{\sqrt{a^{2}-1}}{a} \\
\Rightarrow a=4 .
\end{array}
$$

Therefore, the point $M(-4,0)$.