Problem 6. Point $A$ on the plane is located at the same distance from all points of intersection of two parabolas given in the Cartesian coordinate system on the plane by the equations $y=-3 x^{2}+2$ and $x=-4 y^{2}+2$. Find this distance.
Answer: $\frac{\sqrt{697}}{24}$.

Solution. By plotting the parabolas with the given equations, it is easy to notice that they have 4 common points. Therefore, a point equidistant from them can be at most one. Rewrite the equations of the parabolas as $x^{2}+\frac{1}{3} y-\frac{2}{3}=0$ and $y^{2}+\frac{1}{4} x-\frac{1}{2}=0$ and add them. We get the equation

$$
x^{2}+y^{2}+\frac{1}{4} x+\frac{1}{3} y-\frac{7}{6}=0 \Longleftrightarrow\left(x+\frac{1}{8}\right)^{2}+\left(y+\frac{1}{6}\right)^{2}=\frac{697}{576}
$$

This is the equation of a circle with center $\left(-\frac{1}{8} ;-\frac{1}{6}\right)$ and radius $\frac{\sqrt{697}}{24}$, and it satisfies the coordinates of all intersection points of the parabolas, while the center of the circle is equidistant from these intersection points, with a distance equal to the radius of the circle.