I encountered the following problem:
"We are given n points in the plane such that no three are collinear and no four lie on the same circle. How many circles are there that contain three of these points?"
Now I attempted to solve it and I ended up having 3 circles as the answer. I got there by actually drawing possibilities. However, I feel it's wrong. Any correction or explanation is very much appreciated.
The key geometric fact here is that any $3$ non-collinear points determine a unique circle. To see this, note that the center has to be equidistant from each of your three points, hence must lie on each of the three perpendicular bisectors of the segments made by taking any $2$ of the points. Non-collinearity ensures that no two of these perpendicular bisectors are parallel, To see that all three meet at a single point (the center of the circle we seek): consider two bisectors, $P_{A,B}$ and $P_{B,C}$. As they are not parallel they meet at a point $O$. Now, since $O\in P_{A,B}$ we have $d(O,A)=d(O,B)$. Similarly $d(O,B)=d(O,C)$ It follows that $d(O,A)=d(O,C)$ whence we conclude that $O\in P_{A,C}$.
Therefore, we get a circle of the type you want by taking any three distinct points from your collection. No two of these circles coincide (for if they did then we'd have at least $4$ points from your collection on a single circle, contrary to assumption).
Therefore the number of circles of the type you want equals the number of ways we can choose three points from your collection. Hence the answer is $\binom n3$.