Dividing a circle into areas lines (there are two points on one side and one on the other). So the recurrence can be expressed as which can be easily reduced to using the sums of the first formula_3 natural numbers and the first formula_3 squares, this combines to Finally which yields The lemma asserts that the number of regions is maximal if all "inner" intersections of chords are simple (exactly two chords pass through each point of intersection in the interior). This will be the case if the points on the circle are chosen "in general position". Under this assumption of "generic intersection", the number
Circle of a sphere there is exactly one circle that can be drawn through three given points. The proof can be extended to show that the points on a circle are all a common angular distance from one of its poles. To show that a non-trivial intersection of two spheres is a circle, assume (without loss of generality) that one sphere (with radius formula_1) is centered at the origin. Points on this sphere satisfy Also without loss of generality, assume that the second sphere, with radius formula_3, is centered at a point on the positive x-axis, at distance formula_4 from the origin. Its points