See the 1st part of the solution of problem All such circles pass through two fixed points.

Let $O_1 = (x_1, y_1)$, $O_2 =(x_2, y_2)$ , $r_1, r_2$ be the centres and radii of $C_1, C_2$ respectively. A circle of center $B$ and radius $R$ meets the criteria if and only if $$\overline{BO_1}^2+r_1^2=R^2=\overline{BO_2}^2+r_2^2$$by the pythagorean theorem applied to the triangles connecting $B$, $O_i$ and the point of intersection of both circles (it is clear that $B$ lies on the perpendicular bisector of the common chord). This equation simplifies into a linear equation of the form $$2\Delta_xx+2\Delta_yy =\Delta_{x^2}+\Delta_{y^2}-\Delta_{r^2},$$where we write we write $\Delta_z = z_2-z_1$ for short. We can observe that the locus described by this equation is a line perpendicular to the segment $\overline{O_1O_2}$, and that's a good place to stop.