Arc in a finite plane

Remarks

Let $\Pi$ have order $n$, so each point lies on $n + 1$ lines. If $P$ is a point of a $k$-arc $K$, the lines joining $P$ to the other $k - 1$ points of $K$ are distinct. Hence $k - 1 \le n + 1$, that is, $k \le n + 2$. When $n$ is odd, $k \le n + 1$. An arc with $n + 1$ points is called an oval, and one with $n + 2$ points a hyperoval.

Example: in the plane $\mathrm{PG}(2, q)$, let $K$ consist of the points $(1 : t : t^2)$ for $t \in \F_q$ and the point $(0 : 0 : 1)$. Three points $(1 : s : s^2)$, $(1 : t : t^2)$, $(1 : u : u^2)$ with distinct $s, t, u$ are not collinear, because their Vandermonde determinant $(t - s)(u - s)(u - t)$ is nonzero. For $(1 : s : s^2)$, $(1 : t : t^2)$ and $(0 : 0 : 1)$ the determinant is $t - s \ne 0$. So $K$ is a $(q + 1)$-arc.

Non-example: the $n + 1$ points of a line, for $n \ge 2$, contain three collinear points, so they do not form an arc.

More generally, a $(k, d)$-arc is a set of $k$ points with at most $d$ on any line, so that a $k$-arc is a $(k, 2)$-arc.