IS There a proof that if 4 points on a parabola have x coordinates summing to 0 then there is a circle that contains them all? So I observed that it seems like if the x coordinates of 4 points on a parabola sum to 0 they all lie on a circle, is this true and is there any way to prove it? If so what is the proof?  Thanks!
 A: Consider the parabola $$y=x^2$$
Consider an arbitrary circle with midpoint $(a,b)$ and radius $r$. The corresponding equation is $$(x-a)^2+(y-b)^2=r^2$$
The intersections $(x,y)$ of the circle with the parabola thus have $$r^2=(x-a)^2+(x^2-b)^2=x^4+(1-2b)x^2+(-2a)x+(a^2+b^2)\\x^4+(1-2b)x^2+(-2a)x+(a^2+b^2-r^2)=0$$
If the circle intersects at four points $(x_1,y_1),...,(x_4,y_4)$, this polynomial will have four roots $x_1,...,x_4$, so it equals $(x-x_1)...(x-x_4)$. The $x^3$-coefficient of this will be $-x_1-...-x_4$. So we find that the sum of the $x$-coordinates of the four intersections will have to be zero.
Note that the existence of three different real roots of the polynomial implies the existence of a fourth root, although you might find a root of higher order.
Now if you take any three points on the parabola, there is a unique circle through all three points. The corresponding parabola thus has three different real roots, and thus a fourth, and the sum of the four equals zero.
Thus, if you take four arbitrary points on the parabola with sum of $x$-coordinates zero, the first will lie on the unique circle through the other three, so they all lie on a circle.
A: Consider the parabola $y=ax^2$. Then your statement is equivalent to the existence of a point $(x’,y’): (x_i-x’)^2+(y_i-y’)^2=c, \forall{}i\in\{1,2,3,4\}$ where $c$ is fixed. This is equivalent to $x_i$ being the roots of some polynomial $ P(x)=a^2x^4+(1-2y’a)x^2-2x’x-c’$ for some $x’,y’,c’\in\mathbb{R}$ (here $c’={x’}^2+{y’}^2-c$). This is obviously possible since $1-2y’a,2x’,c’$ are independent values and the coefficient of $x^3$ is always $\sum{}x_i=0$. 
A: The result is true, at least for the parabola $4ay=x^2$. Here is a thoroughly unenlightening algebraic proof.
Let the parabola be parametrised by $x=2at$, $y=at^2$. Then if we consider the curve as lying in the complex plane, it takes the form $ z = x+iy = 2at + iat^2 $. We may then take the four points as $z_k = 2at_k + iat_k^2$ for $k \in \{1,2,3,4\}$, with .
Four points in the complex plane lie on a circle if and only if their cross-ratio is real, that is, if
$$ \Im \left( \frac{(z_1-z_2)(z_3-z_4)}{(z_1-z_4)(z_2-z_3)} \right) = 0 . $$
A tedious computation reveals that the left-hand side of this equation is
$$ \frac{2(t_1-t_2)(t_1-t_3)(t_2-t_4)(t_3-t_4)(t_1+t_2+t_3+t_4)}{(t_2-t_3)(t_1-t_4)(4+(x_2+x_3)^2)(4+(x_1+x_4)^2)} , $$
which is zero if $t_1+t_2+t_3+t_4=0$, as required.

More persuasively, perhaps, we want to find a circle through the four points $(2at_k,at_k^2)$. Such a circle has the form
$$ (x-A)^2 + (y-B)^2 = r^2 , \tag{1} $$
and we want expressions for $A,B,r$ so that $ (2at_k,at_k^2) $ satisfy this; i.e. they are the four roots of the equation
$$ (2at-A)^2 + (at^2-B)^2  - r^2 = 0 , $$
which we can multiply out to find
$$ t^4 + \left(4-2\frac{B}{a}\right)t^2 - 4\frac{A}{a} t + \frac{A^2+B^2-r^2}{a^2} = 0 . $$
Using Vieta's formulae, the roots of this equation satisfy
$$ t_1 + t_2 + t_3 + t_4 = 0 \\
\sum_{i<j} t_i t_j = 4-2\frac{B}{a} \\
\sum_{i<j<k} t_i t_j t_k = 4\frac{A}{a} \\
t_1 t_2 t_3 t_4 = \frac{A^2+B^2-r^2}{a^2} , $$
which gives expressions for $A,B,r$ in terms of the $t_k$. It's not obvious that $r^2>0$ with this approach, but it turns out that if the condition on the $t_k$ is satisfied, it is a product of positive terms. Hence we find real $A,B,r$ so that the four points lie on the circle $(1)$.
