They are actually all tangent to two fixed circles. If $M$ is the intersection of $\ell$ and the perpendicular bisector of $AB$, and the circle centered at $M$ passing through $A, B$ intersects $\ell$ at $T_A, T_B$, then these two fixed circles are the circles centered at $O_A, O_B$ on line $AB$ and tangent to $\ell$ at $T_A, T_B$. To show that $(OCD)$ is always tangent to $(O_A), (O_B)$, it suffices to show that the inverses of $(O_A), (O_B)$ wrt $(ABCD)$ are also tangent to $\ell$. Thus, it suffices to show that $(O_A), (O_B)$ are orthogonal to $(ABCD)$, which is equivalent to $(ABCD)$ being fixed under inversion about $(O_A), (O_B)$. However, by Power of a Point from $O_A$ to $(ABCD)$, $O_AT_A^2 = O_AA \cdot O_AB$, so $(O_A)$ and $(ABCD)$ are orthogonal, and we are done. $\square$