Is there a geometric construction that finds the intersection of two conics, given only five points on each?

Consider the following two sets of points: $$\{(-2,4),(-1,1),(0,0),(1,1),(2,4)\}$$ $$\{(2,2),(2,-2),(4,0),(4/5,8/5),(4/5,-8/5)\}$$ All these points are rational, so there is nothing that cannot be constructed by compass and straightedge alone that could potentially make the unconstructible constructible. The first set defines the parabola $y=x^2$ and the second defines the circle $(x-2)^2+y^2=4$. They meet in two places; one is the origin $(0,0)$ and the other is $$\left(\alpha-\frac1{3\alpha},\alpha^2-\frac23+\frac1{9\alpha^2}\right)$$ where $\alpha=\sqrt[3]{2+\frac{\sqrt{327}}9}$. The minimal polynomials of the coordinates of the second point of intersection are cubic, which means this second intersection cannot be constructed with compass and straightedge alone.

Thus the desired construction does not exist in general. Finding the coefficients of the implicit equations of the conics is possible, however, since it only requires basic linear algebra.

As a simpler counterexample, consider the parabola $y=x^2$ and the circle $y^2=x(1-x)$. Both have infinitely many constructible points and they intersect at the origin. But the second point of intersection requires solving $x^4=x(1-x)$ with $x$ nonzero, leading to the irreducible equation $x^3+x-1=0$ whose real root cannot be (Euclidean) constructed.