Yes, I can.

Let's add additional constraints I've mentioned above: we don't have three points lying on the same straight line and we don't have four points lying on the same circle.

Let's first solve a very easy case of our problem: we have only 3 points. Obviously, for obtuse triangle the center of the enclosing circle will be located at the center of the largest side, otherwise it'll coincide with the center of the circumcircle.

Let's formulate the problem in another terms: we have a function f(x, y), which equals to the distance between (x, y) and the third nearest point to it. We need to find the global minimum of this function.

Let's notice a fact that for arbitrary (x, y) three nearest points to it lie in Voronoi's cells that form a connected area. It can be easily proved by contradiction. But this cells not only form a connected area, but are pairwise adjacent, due to our constraints (it's pretty obvious, but if you need a proof — I'll leave it as a task for you).

But since for each (x, y) three nearest points lie in adjacent Voronoi's cells, these points belong to a triangle of the Delaunay triangulation. Q.E.D.

UPD. I've made a huge mistake. It's not true that the cells of the three nearest points are pairwise adjacent. So, the proof and the solution are incorrect and we need to find how to deal with the problem I've found. My best solution now is again O(N²)

Counterexample:

6
-2 0
2 0
-5 1
5 -1
0 8
0 -8

Ahh, ok. I think I should read the question first :) I assumed by the tittle that it was the standard 'Minimum Enclosing Circle' problem