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Source statement and compiled scope. Mathialagan, published 2021 PDF, pp. 16--17, Lemma 34, states that if ∣A∣≥2|A|\geq2 and BB consists of xx points of a one-dimensional algebraic variety of degree DD, then D(A,B)=ΩD(x)D(A,B)=\Omega_D(x). The source proves this with its preceding algebraic geometry inputs.

Only a line or a circle is needed by Theorem 3. We compile these cases completely, with the explicit bound

∣B∣≤2D(A,B).(1)|B|\leq 2D(A,B). \tag{1}

The unused general algebraic-variety assertion and its external Bézout/component-counting dependencies are not claimed as compiled here.

Proof for a line. Choose any a∈Aa\in A. A circle centered at aa of positive radius intersects a line in at most two points: parameterizing the line and imposing the squared radius gives a nonconstant quadratic equation with positive leading coefficient. Radius zero contributes at most one point. Thus each distance from aa accounts for at most two points of BB, which proves (1).

Proof for a circle. Let the containing circle have center qq and positive radius rr. Since AA contains two distinct points, choose a∈A∖{q}a\in A\setminus\{q\}. A circle centered at aa and the circle centered at qq have at most two common points: subtracting their squared equations gives a genuine line because their centers differ, and the preceding quadratic argument applies. Again distance zero contributes at most one point. Every distance from aa accounts for at most two elements of BB.

Application. With A=PA=P and B=Q∩γB=Q\cap\gamma, where γ\gamma is any line or circle, (1) gives ∣Q∩γ∣≤2D(P,Q)|Q\cap\gamma|\leq2D(P,Q). This includes overlapping P,QP,Q. A circle of radius zero contains at most one point and is never needed in the regulus constructions.

Verification scope. Verified within the independently reviewed Theorem 3 chain, retained in the final review; the exact line/circle specialization and its use in Lemma 26 belong to the living Theorem 3 record.

Bears on. Problem 661.