Set of points in the plane which is intersected by every line on the plane and in which no more than K points are collinear

Question

• Question Let $K \in \mathbb{N} (K \geq 3)$ and $r \in \mathbb{R}^+$. Either find a set $S$ of points in the plane such that every line on the plane intersects atleast one point in $S$ and that no more than $K$ points in $S$ are collinear, or prove that there is no such set.

• *As an additional constraint the points should be so distributed that every circle disc of radius r on the plane has at-least one point which belongs to S*

References on related results would also be nice.

Remark: The question is a kind of non-connected version of this question " find a curve which intersects all lines on the plane at most $K$ times and at least once."

In a more general setting, we may replace the 'circledisc' constraint by a chosen statistical distribution of points.

EDIT

By finding it would be correct to mean a constructive proof

Answer 1

Clearly $K$ must be at least $2$. Under AC (the Axiom of Choice), $K=2$ can be attained, even if we require $S$ to meet every circle, not just circles of fixed radius. The construction uses transfinite induction, so "finds" $S$ only in a somewhat weak sense...

The set of lines and circles in the plane, call it $\Sigma$, has cardinality $c$ (continuum). Using AC we can well-order $\Sigma$ so for each $\alpha \in \Sigma$ there are fewer than $c$ lines and circles preceding $\alpha$ in the order. We now construct $S = \{ p_\alpha : \alpha \in \Sigma \}$, where each $p_\alpha \in \alpha$ is chosen inductively so that it is not collinear with $p_\beta$ and $p_\gamma$ for any distinct $\beta,\gamma \prec \alpha$. This is possible because there are $c$ points in $\alpha$ but the cardinality of lines $\overline{p_\beta p_\gamma}$ with $\beta,\gamma \prec \alpha$ is less than $c$ (if a set has cardinality less than $c$ then so does its square), and each line meets $\alpha$ in at most two points. This fails only if $\alpha$ happens to be the line joining some $p_\beta$ and $p_\gamma$, but then $\alpha$ already has a point of $S$ so we can skip $\alpha$ (or declare that $p_\alpha = p_\beta$). Then $S$ meets every line and every circle, and contains no three collinear points.

The same trick applies in greater generality; e.g. $S$ can be made to meet every algebraic curve and have neither three collinear nor four concyclic points.

(All this must be known already, but it's easier to write a proof than find a reference.)

[Added later I see now that James Cranch suggested this approach in his comment on this question on mathoverflow before it was migrated here.]