The comment by Narasimham made me aware of a very elegant way of tackling this problem. The figure above can be interpreted as the orthogonal projection of a right cone whose axis lies in the plane. The cone intersects the plane in the two lines, $g$ and $h$. The points $A,B,C$ are in fact points on the cone, so they lie either above the plane or below the plane but are projected orthogonally into the plane. These three points in space define a plane, and that plane intersects the cone in a conic section. The orthogonal projection of that conic section is again a conic section, namely one of the four indicated in the figure. The four different solutions come from different choices about which of the points $A,B,C$ lie above the plane and which below. Since reflecting everything in the plane doesn't affect the resulting projected conic, one of the three points can be chosen arbitrarily, while the other two each allow for two possible choices, leading to $2^2=4$ generally distinct solutions.
So let's make this a bit more explicit. Using a suitable projective transformation defined by four points and their images, one can achieve a situation where the lines $g$ and $h$ intersect in the point $(0:0:1)$, the line $g$ intersects the line $AB$ in $(1:0:0)$ and the line $h$ intersects the line $AB$ in $(0:1:0)$. Furthermore, $C=(1:1:1)$ can be the fourth point defining this transformation. Then $A=(a:1:0)$ and $B=(b:1:0)$ describe the situation up to that projective transformation, so we only have to deal with two parameters $a,b\in\mathbb R$ except for some degenerate situations (like when $A$ or $B$ lies on $h$).
Now lift everything up to the cone. That cone has an aperture of $\frac\pi2$. In affine coordinates, you can describe it as the set of points $(x,y,z)$ which satisfies $(x + y)^2 = x^2 + y^2 + z^2$ or in other words $2xy = z^2$. But we are free to scale the $z$ coordinate by $\sqrt2$ so we might as well use
$$xy=z^2\tag1$$
as the equation of the cone. That equation is already homogeneous, so we can plug coordinates $(x:y:z:w)$ into that and find that $w$ is irrelevant. Translating out 2d points above to 3d we obtain $A=(a:1:\pm\sqrt a:0)$ and $B=(b:1:\pm\sqrt b:0)$ as well as $C=(1:1:1:1)$. The plane spanned by these three points is characterized by
$$\begin{vmatrix}1&\pm\sqrt a&0\\1&\pm\sqrt b&0\\1&1&1\end{vmatrix}x
-\begin{vmatrix}a&\pm\sqrt a&0\\b&\pm\sqrt b&0\\1&1&1\end{vmatrix}y
+\begin{vmatrix}a&1&0\\b&1&0\\1&1&1\end{vmatrix}z
-\begin{vmatrix}a&1&\pm\sqrt a\\b&1&\pm\sqrt b\\1&1&1\end{vmatrix}w
=0\tag2$$
If we introduce new symbols $p_i$ for the coefficients of this plane, we can shorten this to
\begin{align*}
p_1x + p_2y + p_3z + p_4w &= 0 \\
p_1x + p_2y + p_4w &= -p_3z \\
(p_1x + p_2y + p_4w)^2 &= p_3^2xy \tag3
\end{align*}
This is a homogeneous quadratic equation in $(x:y:w)$ and as such describes a conic in the original plane. Now one might want to undo the projective transformation which led to the special coordinates, and then we are done. The four possible choices for the signs of $\pm\sqrt a$ and $\pm\sqrt b$ will lead to the four possible conics.
PERHAPS THIS HELPS
I've been thinking about your problem for a while. I wouldn't say that I could solve it. However, I have an idea that I would like to share with you.
You certainly remember the Dandelin spheres which may have something to do with your conjecture. The following figure depicts two cones (a black one and a grey one). These cones share the vertical axis and a Dandelin sphere. Tangent to this sphere there is a blue plane whose edge can be seen in the figure:
The blue plane and the black cone intersect in a parabola. Also, the blue plane intersects with the other cone in a hyperbola. The parabola's focus and one of the foci of the hyperbola are common.
My conjecture is that this is how to imagine your couple of parabola and hyperbola with a common focus.
Then, I don't know how to proceed.
Best Answer
I think it just works for the same reasons you suggested, but some careful backwards implications.
We know we can always find a circle passing through $3$ points. Moreover, given $3$ out of $4$ points whose $x$-coordinates sum to $0$, the fourth point is fixed. So it suffices to show that the circle passing through these $3$ points also happens to pass through this unique fourth point. This is equivalent to showing that the intersection points of any circle (as the circumcircle of those $3$ points is any circle) with $y=x^2$ have a sum of $x$-coordinates that is $0$.
Let's say the circle is $$x^2+y^2+ax+by+c=0$$ Note that its intersections with $y=x^2$ can be found as the roots to the quartic $$x^2+x^4+ax+bx^2+c=0$$ Note that this has no coefficient of $x^3$, so the sum of its roots are $0$.