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For a finite planar point set $P$, its distinct-distance set is $\Delta (P)=\{|p-q|:p,q\in P\}$. It includes zero for nonempty $P$. Its cardinality measures how many different distances the configuration determines. Ordered point pairs can be grouped according to their distance, which relates the problem to incidences between points and curves.

Let $t$ be the maximum number of positive Euclidean distances from one of the $n$ points to the others. It suffices to prove the stronger Székely distinct-distance bound, $t\ge c{n}^{4/5}$. Here $n\ge 2$; enlarge constants to handle bounded $n$.

First prove the crossing lemma for multigraphs. In an optimal drawing of a loopless multigraph with $v$ vertices, $e$ edges and maximum parallel multiplicity $\mu$, crossings of incident edges can be removed by interchanging initial segments, so each remaining crossing involves four distinct vertices. In each parallel class choose one of its edges with probability $1/\mu$ per edge, selecting none with the remaining probability. Independently keep each vertex with probability $p$. The resulting simple graph has expected edge count $p^{2}e/\mu$, expected vertex count $pv$, and expected crossings $p^{4}X/{\mu }^2$, where $X$ is the original crossing count. Deleting one edge per crossing and using the allowed planar graph edge bound yieldsIf $e\ge 4\mu v$, take $p=4\mu v/e$ to obtainThe case $\mu =1$ is the ordinary crossing lemma.

We also need the rich line bound. Given $L$ lines, connect consecutive incidence points on each line by straight edges. This simple graph has at least $I-L$ edges, where $I$ is the total number of incidences, and at most $L^2$ crossings. The crossing lemma gives $I=O((nL{)}^{2/3}+n+L)$, proving the Szemerédi–Trotter theorem here rather than assuming it. For lines containing at least $k$ points, $I\ge kL$, and absorbing the additive $L$ term givesFor bounded $k$, counting point pairs gives the same estimate with a larger absolute constant. Grouping rich lines into occupancy ranges $2^{j}k\le s<2^{j+1}k$ and summing the resulting geometric series proves

Now form the circular-arc graph for distinct distances. Around every point draw each positive-distance circle centred there; there are at most $nt$ such circles. A circle with at least three of the original points contributes one edge between each consecutive pair around it. Discard the circles with at most two points. Since the total number of centre-to-point incidences is $n(n-1)$, the remaining loopless multigraph hasIts circular-arc drawing has $O(n^2{t}^2)$ crossings: two distinct circles have at most two intersections, and small perturbations remove tangencies and multiple crossings without changing this order.

The apparent difficulty is unbounded parallel multiplicity. The rich-bisector deletion bound controls it. Every centre supporting an arc between fixed endpoints lies on their perpendicular bisector. Distinct parallel arcs come from distinct centres, since their common endpoints determine the radius once the centre is fixed. Thus multiplicity at least $k$ makes that line contain at least $k$ of the points. For a line with $s$ centres, each of its at most $st$ circles has at most two consecutive-point arcs symmetric about that line, one at each end of its diameter along the line. The occupancy sum above therefore bounds the number of high-multiplicity arcs by

If $t\ge n^{4/5}$ there is nothing to prove. Otherwise $t\log n=o(n)$ and $t=o(n)$. Choose $k=\lceil K\sqrt{t}\rceil$ with a sufficiently large absolute constant $K$. The first deletion term is at most a small fixed fraction of $n^2$ and the second is $o(n^2)$. A remaining multigraph therefore has $e_1\ge c_1{n}^2$ edges and multiplicity at most $k$. For large $n$, $e_1\ge 4kn$, so the crossing lemma for multigraphs givesRearranging,The whole distinct-distance set contains those distances from the selected point, proving the requested bound. The circular-arc method and the stronger one-point conclusion are due to Székely.

(A) For a finite point set $P$ and a family $\mathcal{C}$ of distinct unit circles in ${\mathbb{R}}^2$, the Szemerédi–Trotter theorem for unit circles stateswith an absolute constant $C$. Here $I$ counts point-circle incidences. Distinctness matters: repeated copies of the same circle are not separate members of this geometric family. A dilation gives the same bound for circles of any one fixed positive radius, with the same constant.

(B) Write $N=|P|$ and $D=|\Delta (P)\setminus \{0\}|$. The comparison is a bound on the cardinality of the distinct-distance set, rather than on the set itself. For each positive distance $\rho$, take the $N$ circles of radius $\rho$ centred at points of $P$. Their incidences between points and curves count exactly the ordered pairs $(p,q)$ at distance $\rho$. After dilation by $1/\rho$, part (A) bounds this number byEvery ordered pair of distinct points contributes to exactly one of these counts. Thus the unit-circle method for a distinct-distance lower bound givesFor $N\ge 2$, $N-1\ge N/2$, soFor $N=1$ the distance set is $\{0\}$ and the conclusion holds after adjusting the absolute constant; the empty set causes no difficulty.

(C) A direct incidence bound from two-point multiplicity suffices. Put $M=|\mathcal{L}|=N^2$ and $k_{\gamma }=|P\cap \gamma |$. Count unordered pairs of distinct points on each curve. By double counting,Writing $I={\sum }_{\gamma }{k}_{\gamma }$, this givesThe Cauchy-Schwarz inequality now yieldsIf $x^2\le ax+b$ with $x,a,b\ge 0$, then $x\le a+\sqrt{b}$. ConsequentlyIn fact the argument proves the stronger $O(N^{9/4})$ bound. The two-point multiplicity hypothesis alone controls these incidences between points and curves; the algebraic degree bound is not needed for the requested estimate.

For a finite set of at least two points in the plane, some point determines the displayed number of different positive Euclidean distances. The circular-arc graph for distinct distances has quadratically many edges. The rich-bisector deletion bound removes high-multiplicity edges while retaining that order of growth. The crossing lemma for multigraphs then compares its lower crossing bound with the quadratic count of pairs of circles. This strengthens the corresponding bound for the whole distinct-distance set. The original result is Theorem11 of Székely's 1997 paper.