$2$ conics intersect at $2$ real points, geometric construction on $\Bbb R^2$ of the line through 2 non-real points?

Question

On $\Bbb R^2$ suppose two conics intersect at two points, then they intersect at two other non-real points on $\Bbb C^2$, it is proved in the comment that the line through the two non-real points is real.

Since this line is real, can we construct this line geometrically on $\Bbb R^2$?

Answer 1

Suppose two conics $a,b$ intersect at two points $P_1,P_2$

Through $P_1$ draw an arbitrary line intersecting $a,b$ at $A_1,B_1$.
Through $P_2$ draw an arbitrary line intersecting $a,b$ at $A_2,B_2$.
Let $C$ be the intersection of the lines $A_1A_2,B_1B_2$.
$C$ is on the desired line, by three conics theorem.

Repeat the construction:
Through $P_1$ draw an arbitrary line intersecting $a,b$ at $A_1',B_1'$.
Through $P_2$ draw an arbitrary line intersecting $a,b$ at $A_2',B_2'$.
Let $C'$ be the intersection of the lines $A_1'A_2',B_1'B_2'$.
$C'$ is also on the desired line.
Then the line $CC'$ is the desired line.

The images are screenshots of Geogebra.
The equation of the two ellipses in the above images are $\cases{a:\;x^2+4y^2=1\cr b:\;x^2-x y + 4y^2 - x = 1}$
They intersect at $P_1(0, -\frac12)$ and $P_2(0,\frac12)$ and non-real points $(\pm i\sqrt3,-1)$.
In Geogebra I verified the line $CC'$ is $y=-1$.

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We can also apply this method if $P_2$ is a point of infinity. For example,
$\cases{a:\;x^2=y-1\cr b:\;x y=1}$ intersect at $P_1,P_2$ and $P_1\in\mathbb R^2$.
$P_2$ is a point of infinity, in homogeneous coordinates $P_2=[0:1:0]$. A line passes through $P_2$ if and only if it is parallel to the $y$-axis.
Using the same method, we can construct the line $CC'$ which passes through the two non-real intersections. By Geogebra, the equation of the line $CC'$ is approximately $2.49102x + 3.65076y = -1.69969$
About the non-zero intersections, their x-coordinates are complex roots of $x^3+x-1=0$, their y-coordinates are complex roots of $y^3-y^2-1=0$.