Baricenter of $4$ intersection points of parabola with circle lies on axis of parabola Show that the baricenter of the $4$ intersection points of a parabola with a circle is on the axis of the parabola.
Let $p$ be a parabola, $c$ a circle and $p\cap c=\{P_1,P_2,P_3,P_4\} \Rightarrow B= \frac{P_1+P_2+P_3+P_4}{4} \in$ axis of $p$.
 A: Without loss of generality we can take the parabola as having equation $y=kx^2$.
Write down the equation $(x-a)^2+(y-b)^2=r^2$ of the circle. Substitute $kx^2$ for $y$.
You will get a degree $4$ polynomial in $x$ with no $x^3$ term. 
But the sum of the roots of the polynomial is  the negative of the coefficient of $x^3$, divided by the coefficient of $x^4$.
So the sum of the four (not necessarily real) roots is $0$. The $x$-coordinate of the barycenter of the points of intersection is, if there are four  real roots, not necessarily distinct, the mean of these $4$ roots.    
A: If we expand a fourth-degree polynomial function $x\mapsto(x-p)(x-q)(x-r)(x-s)$, we get
$$
x^4 - (p+q+r+s)x^3 + \cdots.
$$
The sum of the roots is minus the coefficient of $x^3$.
The $x$-coordinates of the intersection of the parabola $y=x^2$ with the circle $(x-h)^2+(y-k)^2 = r^2$ are the roots of
$$
(x-h)^2 + (x^2-k)^2 = r^2.
$$
That is a fourth-degree polynomial in $x$.  The coefficient of $x^3$ is $0$.  Therefore the sum of the roots is $0$.
