Given two circle $M$ and $N$, we say that $M$ bisects $N$ if they intersect in two points and the common chord is a diameter of $N$. Consider two fixed non-concentric circles $C_1$ and $C_2$.
a) Show that there exists infinitely many circles $B$ such that $B$ bisects both $C_1$ and $C_2$.
b) Find the locus of the centres of such circles $B$.
1. **Given Data and Definitions:**
   - Let \( O_1 = (x_1, y_1) \) and \( O_2 = (x_2, y_2) \) be the centers of circles \( C_1 \) and \( C_2 \) respectively.
   - Let \( r_1 \) and \( r_2 \) be the radii of circles \( C_1 \) and \( C_2 \) respectively.
   - Let \( B \) be the center of a circle that bisects both \( C_1 \) and \( C_2 \).

2. **Condition for Bisecting:**
   - Circle \( B \) bisects \( C_1 \) and \( C_2 \) if the common chord of intersection is a diameter of both \( C_1 \) and \( C_2 \).
   - This implies that the distance from \( B \) to the center of each circle plus the radius of that circle must equal the radius of circle \( B \).

3. **Mathematical Formulation:**
   - For circle \( B \) to bisect \( C_1 \) and \( C_2 \), the following must hold:
     \[
     \overline{BO_1}^2 + r_1^2 = R^2 = \overline{BO_2}^2 + r_2^2
     \]
   - Here, \( \overline{BO_1} \) and \( \overline{BO_2} \) are the distances from \( B \) to \( O_1 \) and \( O_2 \) respectively, and \( R \) is the radius of circle \( B \).

4. **Distance Formulation:**
   - Using the distance formula, we have:
     \[
     \overline{BO_1}^2 = (x - x_1)^2 + (y - y_1)^2
     \]
     \[
     \overline{BO_2}^2 = (x - x_2)^2 + (y - y_2)^2
     \]

5. **Equating the Distances:**
   - Substituting these into the condition, we get:
     \[
     (x - x_1)^2 + (y - y_1)^2 + r_1^2 = (x - x_2)^2 + (y - y_2)^2 + r_2^2
     \]

6. **Simplifying the Equation:**
   - Expanding and simplifying, we obtain:
     \[
     x^2 - 2x x_1 + x_1^2 + y^2 - 2y y_1 + y_1^2 + r_1^2 = x^2 - 2x x_2 + x_2^2 + y^2 - 2y y_2 + y_2^2 + r_2^2
     \]
     \[
     -2x x_1 + x_1^2 - 2y y_1 + y_1^2 + r_1^2 = -2x x_2 + x_2^2 - 2y y_2 + y_2^2 + r_2^2
     \]
     \[
     -2x (x_1 - x_2) + x_1^2 - x_2^2 - 2y (y_1 - y_2) + y_1^2 - y_2^2 + r_1^2 - r_2^2 = 0
     \]

7. **Linear Equation Form:**
   - Rearranging, we get:
     \[
     2(x_2 - x_1)x + 2(y_2 - y_1)y = x_2^2 - x_1^2 + y_2^2 - y_1^2 + r_2^2 - r_1^2
     \]
   - Let \( \Delta_x = x_2 - x_1 \), \( \Delta_y = y_2 - y_1 \), \( \Delta_{x^2} = x_2^2 - x_1^2 \), \( \Delta_{y^2} = y_2^2 - y_1^2 \), and \( \Delta_{r^2} = r_2^2 - r_1^2 \). The equation becomes:
     \[
     2\Delta_x x + 2\Delta_y y = \Delta_{x^2} + \Delta_{y^2} + \Delta_{r^2}
     \]

8. **Locus of Centers:**
   - The above equation represents a line in the \( xy \)-plane. This line is perpendicular to the segment \( \overline{O_1O_2} \).

Conclusion:
- There exist infinitely many circles \( B \) that bisect both \( C_1 \) and \( C_2 \).
- The locus of the centers of such circles \( B \) is a line perpendicular to the segment \( \overline{O_1O_2} \).

The final answer is the locus of the centers of such circles \( \boxed{ B } \) is a line perpendicular to the segment \( \overline{O_1O_2} \).