# How many points are needed to define a circumference?

This doubt comes from a combinatorics problem in a textbook, which states:

Consider two strictly parallel lines and seven dots, four of which are over one of them, and three over the other. Three dots are chosen at random: what is the probability that they define a circumference?

It turns out, the solution is $$1-{{4\choose3}+1\over{7\choose3}}$$. That is, all combinations in which the three chosen dots aren't collinear, divided by all combinations of three dots.

My question is: why mustn't they be collinear?

• If three dots are collinear, that means they are on the same straight line. That straight line can't have more than two points in common with any given circle. – Gerry Myerson Mar 20 '19 at 11:45
• isn't a line like a circle with infinite radius – Sik Feng Cheong Mar 20 '19 at 11:46
• @SikFengCheong There are contexts where you would consider lines to be infinite-radius circles (like when using the flat plane as a model for hyperbolic or projective geometry), but this is not one of them. – Arthur Mar 20 '19 at 11:49
• The numerator in the "solution," $1-{4\choose3}+1$, is negative. I think you meant $$1-{{4\choose3}+1\over{7\choose3}}$$ – Barry Cipra Mar 20 '19 at 12:42
• Have you already seen collinear points on a circle ? – Yves Daoust Mar 20 '19 at 14:38

If three collinear points $$ABC$$ belong to a circle, then there exists a point $$O$$ such that $$AO=BO=CO.$$
Let's show this is not possible. From the above equality it follows that $$OAB$$ and $$OAC$$ are two isosceles triangles, hence $$\angle OAB=\angle OBA =\angle OCB.$$ But then, in triangle $$OBC$$ we have an external angle $$\angle OBA$$ equal to internal angle $$\angle OCB$$, and that is impossible by Euclid's exterior angle theorem, QED.