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Source statement and compiled scope. Mathialagan, published 2021 PDF, pp. 16--17, Lemma 34, states that if and consists of points of a one-dimensional algebraic variety of degree , then . 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
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 circle centered at 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 accounts for at most two points of , which proves (1).
Proof for a circle. Let the containing circle have center and positive radius . Since contains two distinct points, choose . A circle centered at and the circle centered at 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 accounts for at most two elements of .
Application. With and , where is any line or circle, (1) gives . This includes overlapping . 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.