How to determine whether three ellipses have at least one common intersection point or not? How to establish a criterion described in equation so that it is easy to determine whether three ellipses have common intersection area (point) or not?
Update



 A: We need  oracles that 
(a) test for any point ${\bf p}\in{\mathbb R}^2$ and any ellipse $E$ whether ${\bf p}\in {\rm int}(E)$, 
(b) compute for any ellipse $E$ its center ${\bf m}_E$,  
(c) compute for any two ellipses $E_1$ and $E_2$ the set $\partial E_1\cap\partial E_2$, consisting of $0$, $2$, or $4$ points. 
(Degenerate cases will not be considered.)
Let $A$, $B$, $C$ be  the three given ellipses. 
Compute all intersection points of two of the three boundary ellipses $\partial A$, $\partial B$, $\partial C$, using oracle (c).
If $\partial A\cap\partial B$ is empty check whether ${\bf m}_A\in B$ or ${\bf m}_B\in A$. If ${\bf m}_A\in B$ we know that $A\subset B$, so it remains to check the ellipses $A$ and $C$ versus each other. (I leave this to the OP.) – Similarly when ${\bf m}_B\in A$.  
Similarly when $\partial B\cap\partial C$ or $\partial C\cap\partial A$ is empty.
Assume now that any two boundary ellipses $\partial A$, $\partial B$, $\partial C$ intersect. Then we apply the following Lemma and oracle (a):
Lemma. The set $\Omega:={\rm int}(A\cap B\cap C)$ is nonempty iff at least one  intersection point of two boundary ellipses  lies in the interior of the third ellipse.
Proof. Any neighborhood of ${\bf r}\in\partial A\cap\partial B$ contains points of ${\rm int}(A\cap B)$. When ${\bf r}$ lies in ${\rm int}(C)$ the set $\Omega$ is nonempty.
For the proof of the converse consider the following "Gedankenexperiment": We pick a point ${\bf p}\in \Omega$ and draw a ray from ${\bf p}$ to infinity that does not go through any intersection point of two boundary ellipses. This ray will hit a first boundary ellipse, say $\partial A$, in a point ${\bf q}$. The point ${\bf q}$ is still in the interior of both $B$ and $C$. Now walk from ${\bf q}$ along $\partial A$ until you hit for the first time a point of $\partial A\cap\partial B$ or $\partial A\cap\partial C$. Assume this is  the point ${\bf r}\in\partial A\cap\partial B$, say. This point ${\bf r}$ is still in the interior of $C$.
A: Inspired by @ Semiclassical comment, I got one answer:


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*suppose three ellipse are $a$, $b$ and $c$; so there are three combinations if we draw two from the three: $(a,b),(a,c),(b,c)$

*test the number of interaction points between $(a,b), (a,c),\text{ or } (b,c)$, if the number is $0$, there is no common point between the three; if the number is $1$, then if the only point is not in or on the third ellipse,  there is no common point between the three;

*for the cases when interaction points between all the combinations are more than or equal to $2$ ($\leqslant 4$), test if there is any intersection point in/on the third ellipse; if all are no, then they have no common point; if any is yes, then the three ellipses have common interaction points.
This still has exception case:

Update


*

*suppose three ellipse are $a$, $b$ and $c$; so there are three combinations if we draw two from the three: $(a,b),(a,c),(b,c)$

*test the number of interaction points between $(a,b), (a,c),\text{ or } (b,c)$, 
2.1 if the number is $0$, check whether one ellipse is totally contained by another; if no, there is no common point between the three; if yes, the problem becomes determine whether the small ellipse and the third have common point or not; check  the existence of intersection point and the containing relationship between the two;
2.2 if the number is $1$, then if the only point is not in or on the third ellipse,  there is no common point between the three;

*for the cases when interaction points between all the combinations are more than or equal to $2$ ($\leqslant 4$), test if there is any intersection point in/on the third ellipse; if all are no, then they have no common point; if any is yes, then the three ellipses have common interaction points.
