Q:

If a disk contains the diagonal of a quadrilateral, it will contain at least one of the vertices not lying on this diagonal

Final Version:
Given a convex quadrilateral, then at least one of the two diagonals satisfy this: if a disk contains it, then it should contain one of the rest two vertices of the quadrilateral.
2nd Version:
Suppose we have four points in the plane in general position, call them $a,b,c,d$ and take the quadrilateral they form. Take any disk that contains the longest diagonal. Then, this disk will contain at least one of the point not on the longest diagonal.
Edit: Where my question comes from? Stating again a new version of the problem.
I need a result like this for a problem in Machine Learning. The problem comes down to this. Given a circle in the plane, the points inside it are marked as $+1$ and points outside it as $-1$. Take the family of all circles in the plane and consider four arbitrary points in the plane. I want to show that I can't achieve all the possible labellings for these points by using circles from my family -we say in such case that these points can't be shattered. To make this more clear, let me mention for example that if we have three points in the plane then obviously we can take all the $2^3$ possible labelings of the three points by taking a circle that contains all of them, none of them, exactly one of them (x3) and exactly two of them (x3). I know that in the case of four points we can't do so and I am trying to prove that. 
If the convex hull of the four points is a segment, then obviously these points can't be shattered -if they are ordered  $a,b,c,d$ in the segment then the labelling $(1,-1,1,-1)$ can't be realised, since a circle that contains $a$ and $c$ will also contain $b$ as well. 
If the convex hull of the four points is a triangle, then again these points can't be shattered -if $a,b,c$ are the vertices of the triangle then the labelling $(1,1,1,-1)$ can't be realized.
So we are left with the case that the convex hull is a quadrilateral. In this case, it should be that there exists a labelling that can't be realized. If our quadrilateral is $a,b,c,d$ then I can see that all of the labellings can be realised except for $(1,-1,1,-1)$ or $(-1,1,-1,1)$. I thought that it would be the case that the labelling that fails to be realised is the one with $1's$ at the longest diagonal, but that's not the case, as Raffaele's answer suggests. 
It seems sufficient to prove the following: given a quadrilateral, then at least one of the two diagonals satisfy this: if a disk contains it, then it should contain one of the rest two vertices of the quadrilateral.
Is this now true? 
First Version:
Suppose we have four points in the plane in general position, call them $a,b,c,d$. Also suppose that $a,b$ have the longest distance among these points. Take any disk that contains this segment. Then, this disk will contain at least one of $c,d$.
It seems obvious intuitively that this holds, but I failed to prove it rigorously.
Any suggestion?

A:

Given a convex quadrilateral, then at least one of the two diagonals satisfy this: if a disk contains it, then it should contain one of the rest two vertices of the quadrilateral.

I am glad to give a happy end to this question story. The answer is positive. Indeed, denote the vertices of the quadrilateral (in cyclic order) by $A$, $B$, $C$, $D$. Assume that a disk contains a diagonal $AC$, but none of the vertices $B$  and $D$. Let $A’C’$ be a chord of the disk containing the side $AC$. Then $\angle ABC+\angle ADC\le \angle A’BC’+\angle A’DC’<\pi$. Similarly, if an other disk contains a diagonal $BD$, but none of the vertices $A$  and $C$ then $\angle BAD+\angle BCD <\pi$. But then 
$$2\pi=\angle ABC+\angle ADC+\angle BAD+\angle BCD<2\pi,$$
a contradiction.

