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In this question a method is presented for solving a cubic equation using strainghtedge, compasses and a single additional conic section,the parabola $y=x^2$. Briefly, the method starts by introducing an auxiliary root whise sum with the cubic roots is zero, thus the cubic equation becomes a depressed quartic. The quartic equation is then converted to a two-dimensional quadratic equation for a circle using the parabola equation $y=x^2$. When the circle is intersected with the parabola, the $x$-coordinates of the intersection points give the cubic root(s) along with the auxiliary one.

So when the cubic has three real roots there will be four points of intersection, which by this construction have $x$-coordinates summing to zero and also are concyclic.

The question is whether this is more than a coincidence. Given four points on the parabola $y=x^2$, are they always concyclic when the $x$-coordinates sum to zero?

Oscar Lanzi
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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 $0$.

Alan Abraham
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