An ordinary circle of a set P of n points in the plane is defined as a circle that contains exactly three points of P. We show that if P is not contained in a line or a circle, then P spans at least $$n^2/4 - O(n)$$n2/4-O(n) ordinary circles. Moreover, we determine the exact minimum number of ordinary circles for all sufficiently large n and describe all point sets that come close to this minimum. We also consider the circle variant of the orchard problem. We prove that P spans at most $$n^3/24 - O(n^2)$$n3/24-O(n2) circles passing through exactly four points of P. Here we determine the exact maximum and the extremal configurations for all sufficiently large n. These results are based on the following structure theorem. If n is sufficiently large depending on K, and P is a set of n points spanning at most $$Kn^2$$Kn2 ordinary circles, then all but O(K) points of P lie on an algebraic curve of degree at most four. Our proofs rely on a recent result of Green and Tao on ordinary lines, combined with circular inversion and some classical results regarding algebraic curves.

中文翻译：

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关于定义几个普通圆的集合

平面中由 n 个点组成的集合 P 的普通圆被定义为恰好包含 P 的三个点的圆。我们证明，如果 P 不包含在一条直线或一个圆中，则 P 至少跨越 $$n^ 2/4 - O(n)$$n2/4-O(n) 个普通圆。此外，我们为所有足够大的 n 确定了普通圆的确切最小数量，并描述了接近该最小值的所有点集。我们还考虑果园问题的圆形变体。我们证明 P 最多跨越 $$n^3/24 - O(n^2)$$n3/24-O(n2) 个圆，正好通过 P 的四个点。这里我们确定了精确的最大值和极值配置对于所有足够大的 n。这些结果基于以下结构定理。如果 n 根据 K 足够大，并且 P 是一组 n 个点，最多跨越 $$Kn^2$$Kn2 个普通圆，那么 P 的除 O(K) 个点之外的所有点都位于最多为 4 次的代数曲线上。我们的证明依赖于 Green 和 Tao 在普通直线上的最新结果，结合圆形反演和一些关于代数曲线的经典结果。