Understanding tangencies between two ellipses. Let $E_1,E_2$ be two (distinct) ellipses in $\mathbb{R}^2$. The intersection between $E_1$ and $E_2$ may contain from $0$ to $4$ points and one way to determine these points, is to note that they are the solutions of an fourth order polynomial equation: $$P(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4=0. \tag{1}$$
I am interested in understand when $E_1$ and $E_2$ are tangents. Geometrically we have two kind of tangencies, to wit, $E_1$ and $E_2$ are internally tangents and by this I mean that, $E_1$ contains $E_2$ in its interior or $E_2$ contains $E_1$ in its interior; and they are externally tangents, which means that they are not internally tangents.

It seems that both cases corresponds to when $P$ has a real root with multiplicity four; or when $P$ has a real root with multiplicity two and two complex roots (not necessarily the same order as in the previous paragraph).

My questions are:


*

*Is the content of the gray box true? If so, which case correspond to a externally tangencie?

*Does anyone know a way to study the intersection between two ellipses, avoiding the study of equation $(1)$?
The brute force approach is tireless. For example, if $P$ has just one root, let's say, $a$ then, $P(a)=P'(a)=P''(a)=P'''(a)=0$. This is a huge system to study! 
Another brute force approach, is to study the nature  of the roots of $P$, however, $a_i$, $i=0,1,2,3,4$ are huge equations of the coefficients of $E_1,E_2$!
 A: A tangency point corresponds to two real roots coalescing (becoming equal). 
I can't prove it rigorously right now, but my guess is that you'll never get four coincident real roots unless the two ellipses are identical. Here's a fuzzy argument. Given four points, there is a unique ellipse that passes through them. The four-fold real root corresponds to a four-fold point (in some sense). If two ellipses both pass through this same four-fold point, then they must be identical.
A good way to study the intersections is by use of the "pencil" concept. The idea is that if $A(x, y)=0$ and $B(x,y)=0$ are two conic section curves, then, for any given $\lambda$, $C_\lambda(x,y) = A(x, y) + \lambda B(x,y)=0$ is also a conic section curve, and it passes through the intersection points of $A=0$ and $B=0$ (because $A=0$ and $B=0$ implies that $C_\lambda = 0$). As $\lambda$ varies, we get a "pencil" of conics. By a judicious choice of $\lambda$, we can cause $C_\lambda $ to degenerate into a pair of straight lines. Then we just have to intersect these lines with $A$ or $B$, which means you have to solve two second-degree equations.
The "judicious choice of $\lambda$" involves solving a cubic equation, which is obviously a lot easier than solving a quartic. A brief account is given here.
I don't think an analysis of roots will tell you whether the tangency is internal or external. You'll probably have to study the curvatures of the two conics at the tangency point.
If you need code to compute the intersections, you can find it here. I haven't used this specific function, but the code on that site is generally good quality.
