The answers to the question "Circle enclosing all but one of n points" demonstrate that, given $n$ points, it is possible to construct a circle such that all but one of the points is inside the circle (i.e. given a set of points $A_i$, $i=1, 2, \dots, n$, we can find some point $B$ — not necessarily, though possibly, in our set $\{A_i\}$ — and a radius $r$ such that the distance $A_iB < r$ for $n-1$ of the points, while the other point has $A_iB > r$).
Obviously the solution is not unique: there are generally many such ways the circle could be drawn, including by leaving different points outside the circle. But given a set of points $\{A_i\}$, is there a way to identify (preferably by geometric construction rather than a co-ordinate geometry calculation) that subset of points which could be the "odd one out" when a circle enclosing $n-1$ points is drawn?
For example, given any three non-collinear points it seems clear that for any of those points, we could construct a circle that leaves that point alone out. If the three points lie in a line, then it isn't possible to draw a circle that only leaves the middle point out (because circles are convex, and for a convex set then any point on the line segment between two points inside the set must also be inside the set), whereas leaving out either end point is easy. Given four points arranged in a square, it is straightforward to construct circles to leave out any given point; if instead the four points were arranged so that three were the vertices of an equilateral triangle while the fourth is at its centroid, it seems the central point can't be left out but the others can.
My intuition is that any point that does not lie in the convex hull of the set of other points is "leave-outable" because we can then draw a line separating it from the other points, and by using a large enough radius we can construct a circle with a centre on the "$n-1$ points" side of the line, and whose circumference is "close enough" to the relevant segment of this line (sufficiently "approximately straight") for our point to be on the extreme side of the arc just as it was to the line, while making sure the circle is big enough to enclose the desired $n-1$ points. My intuition also suggests that any point that lies in the convex hull of the set of other points is "not leave-outable" for similar reasons to the three collinear points example ... if both these intuitions are correct, a necessary and sufficient condition is "points that don't lie in the convex hull of the other points". Is this intuition correct, and if so, is there a nice proof and/or construction for it?