Oval

Remarks

If $\Pi$ is finite of order $n$, a set $O$ is an oval if and only if $O$ has $n + 1$ points and no three of them on a line. Indeed, suppose $O$ has no three points on a line, and let $P \in O$. Of the $n + 1$ lines through $P$, exactly $\abs{O} - 1$ meet $O$ again. So $P$ has $n + 2 - \abs{O}$ tangents, and this equals $1$ exactly when $\abs{O} = n + 1$.

Example: in $PG(2, q)$ the non-degenerate conic $xz = y^2$ has the $q + 1$ points $[1 : t : t^2]$, $t \in \F_q$, and $[0 : 0 : 1]$. A line $ax + by + cz = 0$, with $(a, b, c) \ne 0$, meets it in the points $[1 : t : t^2]$ with $a + bt + ct^2 = 0$, and in $[0 : 0 : 1]$ exactly when $c = 0$. If $c \ne 0$, the polynomial has at most two roots. If $c = 0$, it has at most one root, and the line also contains $[0 : 0 : 1]$. In both cases the line meets the conic in at most two points. So the conic is a set of $q + 1$ points with no three on a line, hence an oval.

Non-example: three non-collinear points $A, B, C$ in $PG(2, 3)$. No line contains all three, so clause 1 holds. Through $A$ there are $4$ lines, and only $AB$ and $AC$ meet the set again. So $A$ has two tangents, and clause 2 fails.