Given the parabola $y=k x^{2}$. Determine the center of the circle $x^{2}-2 p x+y^{2}-2 q y=0$ such that the abscissas of the intersection points with the given parabola are the roots of the equation $x^{3}+a x+b=0$.
The coordinates of the center $K$ of the given circle are $K(p, q)$. Knowing these is sufficient to draw the circle, as the circle must pass through the origin $O$, since the value pair $x=0, y=0$ satisfies the equation.

The parabola and the circle can have at most 4 common points. Indeed, eliminating $y$ from the two equations results in a fourth-degree equation:

$$
k^{2} x^{4}+(1-2 k q) x^{2}-2 p x=0
$$

Thus, at most 4 real numbers can appear as the abscissas of the common points, and the parabola has only one point for each abscissa. One of their common points is the origin, and the abscissas of their other common points satisfy the equation obtained by factoring out the $x$ factor corresponding to the known $x=0$ root from (1):

$$
k^{2} x^{3}+(1-2 k q) x-2 p=0
$$

Here, $k \neq 0$ - otherwise, instead of a parabola, we would be facing the line $y=0$ - , so the equation is equivalent to the following:

$$
x^{3}+\frac{1-2 k q}{k^{2}} \cdot x-\frac{2 p}{k^{2}}=0
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_218d188fd2da4a5baf2eg-1.jpg?height=827&width=984&top_left_y=828&top_left_x=560)

The abscissas of the common points of the parabola and the circle, different from the origin, give the roots of the equation

$$
x^{3}+a x+b=0
$$

- that is, all real roots of (3) - if (2) is equivalent to (3), or if their other two coefficients are respectively equal:

$$
\begin{array}{rlrl}
\frac{1-2 k q}{k^{2}} & =a, & -\frac{2 p}{k^{2}}=b, \text { and from this } \\
p & =-\frac{k^{2}}{2} \cdot b, & & q=\frac{1}{2 k}-\frac{k}{2} \cdot a
\end{array}
$$

According to these, the real roots of equation (3) can be read off as the abscissas of the common points of the given parabola $y=k x^{2}$ and the circle centered at $K(p, q)$ determined by expressions (4), with radius $K O$. For example, choosing $k=1$, the

$$
K\left(-\frac{b}{2}, \frac{1-a}{2}\right)
$$

circle centered at this point, passing through $O$, must intersect the normal parabola.

In the figure, the solution of the equation $4 x^{3}-13 x-6=0$ is shown, where $p=3 / 4, q=17 / 8$, and $x_{1}=-3 / 2, x_{2}=-1 / 2$, $x_{3}=2$.

András Recski (Budapest, Bolyai J. g. III. o. t.)

Remark. It is natural that in the graphical solution of the cubic equation, the coordinates $p, q$ are also determined graphically. For this purpose, the auxiliary lines in the figure are suitable (still for $k=1$). The equations of these lines are

$$
x=-\frac{y}{2}, \quad \text { or } \quad y=\frac{1-x}{2}
$$

The abscissa $x=-b / 2=p$ corresponds to the point $S_{1}$ with ordinate $y=b$, so the line passing through the auxiliary point $K=S_{1}$ will be on the line parallel to the $Y$-axis. The ordinate $y=(1-a) / 2=q$ corresponds to the point $S_{2}$ with abscissa $x=a$, so the point $K$ is the intersection of the last line with the line passing through $S_{2}$ and parallel to the $X$-axis.