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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.

(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 distinct equal-radius circles and a finite point set in the real plane, incidences between points and curves satisfy $I(P,\mathcal{C})=O(|P{|}^{2/3}|\mathcal{C}{|}^{2/3}+|P|+|\mathcal{C}|)$. Scaling the plane reduces any common positive radius to one. The constant is universal. The fixed-radius condition bounds the number of circles through two prescribed distinct points by two.

For every positive distance, draw equal-radius circles centred at the points of $P$. Their incidences between points and curves count the ordered pairs at that distance. The Szemerédi–Trotter theorem for unit circles bounds each distance class by $O(|P{|}^{4/3})$. Summing over all classes accounts for $|P|(|P|-1)$ pairs, proving $|\Delta (P)|\gtrsim |P{|}^{2/3}$.