I claimed in a comment on @Zander's answer that "a cleaner diagram could even make [his 'Least Algebraic' proof] a Proof Without Words". Well, I've done a cleaning pass for fun, but since the order of construction of the elements isn't necessarily clear, I'll offer a Proof With Words. (Actually, since I tried to reduce labeling clutter in the images, my prose has descriptor clutter, so that this is really a Proof With Way Too Many Words.)
Given parabolas with foci $F$ and $G$ and common chord $AB$ (all marked with stars), let $C$ be the midpoint of $AB$, let $M$ and $N$ be the points where the perpendicular bisector of $AB$ meets the parabolas' axes (as shown), and let $O$ be the point on $MN$ such that $NO = CM$.
In the first image, we drop a perpendicular from $C$, through $P$ on a parabola, to the corresponding directrix. Classic reflection properties ensure both that the tangent line through $P$ is parallel to $AB$, and that it is perpendicular to the hypotenuse of (green) $\triangle F$. Consequently, that hypotenuse is parallel to $MN$, and we have $\triangle F \cong \triangle N \cong \triangle O$; in particular, the height of $\triangle O$ is congruent to the height of $\triangle F$.
Likewise, in the second figure, the perpendicular from $C$ determines $Q$, such that the tangent through $Q$ is parallel to $AB$ and perpendicular to the hypotenuse of (orange) $\triangle G$. Thus, $\triangle G \cong \triangle M \cong \triangle O$, with the width of $\triangle G$ matching the width of $\triangle O$.
The upshot: The displacement of $O$ from the intersection of the parabolas' axes is determined by each parabola's focus-to-directrix distance, and is therefore independent of the choice of common chord $AB$. As $O$ lies on the perpendicular bisector of each such common chord, it is the center of a circle containing the four points of intersection of the two parabolas.
Note. The displacement of $O$ from the axes is independent of the distance from $F$ to the axis through $G$, and independent of the distance from $G$ to the axis through $F$. That is, we can move the "$F$" parabola up and down, and the "$G$" parabola left and right, and $O$ will remain the center of the circle containing the points of intersection. (Of course, the radius of the circle changes.) Interesting.
Note. My figure cheats a little, because I approximated parabolas with ellipses in Photoshop (which really isn't the best tool for this job).