1. **Define the positions of points $A$ and $B$:**
- Point $A$ moves on a circle centered at $O(0,0)$ with radius $r$. Its coordinates are given by:
$$A = (r \cos \beta, r \sin \beta)$$
where $\beta$ is a variable angle.
- Point $B$ moves on a circle centered at $C(-2r,0)$ with radius $r$. Its coordinates are given by:
$$B = (-2r + r \cos(\alpha + \beta), r \sin(\alpha + \beta))$$
where $\alpha$ is a constant angle and $\beta$ is the same variable angle as for point $A$.

2. **Find the midpoint $M$ of segment $AB$:**
- The coordinates of the midpoint $M$ are given by the average of the coordinates of $A$ and $B$:
$$M = \left( \frac{r \cos \beta + (-2r + r \cos(\alpha + \beta))}{2}, \frac{r \sin \beta + r \sin(\alpha + \beta)}{2} \right)$$
Simplifying the coordinates of $M$:
$$M = \left( \frac{r \cos \beta - 2r + r \cos(\alpha + \beta)}{2}, \frac{r \sin \beta + r \sin(\alpha + \beta)}{2} \right)$$
$$M = \left( \frac{r (\cos \beta + \cos(\alpha + \beta)) - 2r}{2}, \frac{r (\sin \beta + \sin(\alpha + \beta))}{2} \right)$$
$$M = \left( \frac{r (\cos \beta + \cos(\alpha + \beta)) - 2r}{2}, \frac{r (\sin \beta + \sin(\alpha + \beta))}{2} \right)$$

3. **Determine the equation of the perpendicular bisector of $AB$:**
- The slope of $AB$ is given by:
$$\text{slope of } AB = \frac{r \sin(\alpha + \beta) - r \sin \beta}{-2r + r \cos(\alpha + \beta) - r \cos \beta}$$
Simplifying the slope:
$$\text{slope of } AB = \frac{r (\sin(\alpha + \beta) - \sin \beta)}{r (\cos(\alpha + \beta) - \cos \beta) - 2r}$$
$$\text{slope of } AB = \frac{\sin(\alpha + \beta) - \sin \beta}{\cos(\alpha + \beta) - \cos \beta - 2}$$
- The slope of the perpendicular bisector is the negative reciprocal of the slope of $AB$:
$$\text{slope of perpendicular bisector} = -\frac{\cos(\alpha + \beta) - \cos \beta - 2}{\sin(\alpha + \beta) - \sin \beta}$$

4. Find the fixed point through which the perpendicular bisector passes:
- The perpendicular bisector of $AB$ always passes through a fixed point $D$. To find this point, we need to determine the coordinates of $D$ such that it satisfies the equation of the perpendicular bisector for all values of $\beta$.
- Given the coordinates of $D$ as $(-r, -r \cot \frac{\alpha}{2})$, we need to verify that this point lies on the perpendicular bisector for all $\beta$.

5. Verification:
- The coordinates of $D$ are $(-r, -r \cot \frac{\alpha}{2})$.
- The perpendicular bisector of $AB$ passes through $D$ if the coordinates of $D$ satisfy the equation of the perpendicular bisector for all $\beta$.
- Since the coordinates of $D$ are independent of $\beta$, $D$ is indeed a fixed point through which the perpendicular bisector of $AB$ passes.

$\blacksquare$