= Solution
\b[(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> states
$$
\boxed{I(P,\mathcal C)\le C\bigl(|P|^{2/3}|\mathcal C|^{2/3}+|P|+|\mathcal C|\bigr)}
$$
with 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[(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 by
$$
C(N^{4/3}+2N)\le C_1N^{4/3}\qquad(N\ge1).
$$
Every ordered pair of distinct points contributes to exactly one of these counts. Thus the <unit-circle method for a distinct-distance lower bound> gives
$$
N(N-1)\le C_1D N^{4/3}.
$$
For $N\ge2$, $N-1\ge N/2$, so
$$
\boxed{|\Delta(P)|\ge D\ge \frac1{2C_1}N^{2/3}.}
$$
For $N=1$ the distance set is $\{0\}$ and the conclusion holds after adjusting the absolute constant; the empty set causes no difficulty.
\b[(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>,
$$
\sum_{\gamma\in\mathcal L}\binom{k_\gamma}{2}
=\sum_{\{p,q\}\subset P}|\{\gamma:p,q\in\gamma\}|
\le\sqrt N\,\binom N2.
$$
Writing $I=\sum_\gamma k_\gamma$, this gives
$$
\sum_\gamma k_\gamma^2
=I+2\sum_\gamma\binom{k_\gamma}{2}
\le I+N^{1/2}N(N-1).
$$
The <Cauchy-Schwarz inequality> now yields
$$
I^2\le M\sum_\gamma k_\gamma^2
\le N^2I+N^{9/2}.
$$
If $x^2\le ax+b$ with $x,a,b\ge0$, then $x\le a+\sqrt b$. Consequently
$$
\boxed{I(P,\mathcal L)\le N^2+N^{9/4}\le2N^{5/2}\qquad(N\ge1).}
$$
In 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.
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