Could You help me give a proof that:
Given a finite number $(>3)$ of points in the Euclidean plane, then exist a circle through three points such that any other points inside (or lie on) the circle.
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Start with the convex hull of the points. Pick any edge of that hull, and extend it to a line. That line can be seen as a degenerate circle of infinite radius, or more usefully of curvature zero. Now start increasing the curvature while maintaining contact with the two incident points. These two points and the curvature uniquely define the circle, which makes this easy to describe. There are two possible scenarios. Either you at some point encounter a curvature for which the circle starts touching another point from your set. In that case you are done.
Or you increase the curvature as much as you can without encountering that situation. In that case you end up with a circle which has the two points antipodal on a diameter, and all the other points within. Which means that by coincidence you picked points which are particularly far apart, farther than any other pair of points in your set (since two points inside the circle have to be less than the diameter away from one another). Start over with a different edge from the convex hull and you are good.
There are some special cases worth considering. It might happen that you already touch a thrid point in the initial configuration, with curvature zero. That can happen if you have collinear points on your convex hull. If you pick the outermost points from such a collinear edge, you do touch the inner points initially, but move away from them as the curvature increases.
If all your points are in a straight line, then choosing the outermost points of each collinear edge leaves you with exactly one pair to choose. You will end up in the position where the circle of maximal curvature still only touches two points, and you won't have a different edge to choose from. So in that case the statement holds only if you also allow circles of zero curvature.
See also A circle with infinite radius is a line.