Counting points of intersection

With $9$ points on the circle, there are $\binom{9}{2}=36$ line segments.

How many couples of line segments share a vertex? For every vertex, there are $8$ line segments with such an endpoint, hence there are $9\cdot\binom{8}{2}=252$ couples of line segments with a common endpoint, and at most:

$$\binom{36}{2}-9\cdot\binom{8}{2} = 378 $$ points of intersection not on the circle. In general, with $n$ points on the circle there are at most: $$ \binom{\binom{n}{2}}{2}-n\cdot\binom{n-1}{2} = \frac{n(n-1)(n-2)(n-3)}{8}=3\binom{n}{4}$$ points of intersection not on the circle. Inside the circle, there are just $\color{red}{\binom{n}{4}}$ points of intersection: every choice of four points on the circle is associated with a unique internal point of intersection. So with $9$ points on the circle there are at most $\color{red}{126}$ internal points of intersection.

You just have to choose where to place these $n$ points in such a way that no triple of line segments is concurring, i.e. you have to prove that you can arrange them in general position to achieve this maximum:

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