General Ellipse Intersection Detection

Question

I'm doing some research on my own to find a method to check if two ellipses intersect.

So let's say we have the fallowing two general equations:

$$ ƒ_{1}\ :\ A_{1}x^{2}\ +\ B_{1}xy\ +\ C_{1}y^{2}\ +\ D_{1}x\ +\ E_{1}y\ +\ F_{1}\ =\ 0 $$

and

$$ ƒ_{2}\ :\ A_{2}x^{2}\ +\ B_{2}xy\ +\ C_{2}y^{2}\ +\ D_{2}x\ +\ E_{2}y\ +\ F_{2}\ =\ 0 $$

Since $k_{1}*0 = k_{2}*0$ any linear combination of the two functions intersects all the other linear combinations of the two (including them two) in the same points they'd intersect like this:

So any linear combination of two intersecting ellipses will be non-imaginary and non-degenerate because it has to pass trough those two real points.

With non-intersecting ellipses instead, some linear combinations "disappear" from the real plan like this:

So since reading this Wikipedia article an ellipse is real if:

$$ (A + C) * det A_{Q} < 0 $$

and given that $det A_{q}$ is:

$$ \left(AC\ -\ \frac{B^{2}}{4}\right)F\ +\ \frac{BED}{4}-\frac{CD^{2}}{4}-\frac{AE^{2}}{4} $$

I thought that, for intersecting ellipses the inequality:

$$ ƒ_{3}\ :\left(\left(\left(A_{1}x+A_{2}y\right)\left(C_{1}x+C_{2}y\right)\ -\ \frac{\left(B_{1}x+B_{2}y\right)^{2}}{4}\right)\left(F_{1}x+F_{2}y\right)\ +\ \frac{\left(B_{1}x+B_{2}y\right)\left(E_{1}x+E_{2}y\right)\left(D_{1}x+D_{2}y\right)}{4}-\frac{\left(C_{1}x+C_{2}y\right)\left(D_{1}x+D_{2}y\right)^{2}}{4}-\frac{\left(A_{1}x+A_{2}y\right)\left(E_{1}x+E_{2}y\right)^{2}}{4}\right)\left(\left(A_{1}x+A_{2}y\right)+\left(C_{1}x+C_{2}y\right)\right)\ <\ 0 $$

Would fill the whole plan since there shouldn't be any possibile combinations of $x, y$ ($k_{1}, k_{2}$) that make the general equation degenerate.

What I got is somehow satisfying but at the same time confusing:

So it seems that if the two ellipses are disjoined (the center of one is outside the other), the first and third quadrant are indeed filled.

Those negative slices in the second and fourth, when the two ellipses are disjoint, overlap with slices where the linear combination is an hyperbole ($A_{33} < 0$):

Since my application would be to detect collision of disjoint entities in a 2D plane, I think this method might work:

1. Determine all coefficients of $ƒ_{3}$
2. Evaluate $ƒ_{3}$ for points with 30° and 60° angle from $y=0$
3. If any of the two is $\le 0$ the two ellipses intersect (or touch)
4. If not, take the lowest result and find points at $\pm 15°$ from it
5. Keep evaluating & splitting the quadrant fallowing the lowest result

This way with like 7 evaluations I can check down to like 0.7° precision.

So here's some question I got:

1. What's the formal demonstration that any linear combination of $ƒ_{1}$ and $ƒ_{2}$ pass for their same intersection points?
2. What are those negative slices in the $ƒ_{3}$ plot?
3. Is there any wrong passage in my reasoning?
4. How "good" would this method be compared to standard methods?
5. Is there any what this can also hint to the actual minimal distance between the two ellipses?

Thanks a lot.