Prove two circles cannot intersect more than twice
Answers
Answered by
0
I am quite ignorant on the topic, but I can't help mention it simply because of its beauty: Algebraic varieties, especially the ones defined on projective spaces. I had to deal with them while trying to prove that a particular set of three cubic polynomial equations in three variables had one and only one real solution. Then, I learned that it was actually about the intersections of algebraic varieties.
A circle is an algebraic variety on the real plane. Two distinct circles will intersect at most at two points, as shown by other answers. However, this is not too satisfactory. Because, the circles sometimes meet at a single point, and sometimes never. Same is true for two lines: they may intersect at one point or they may not intersect at all. We are considering only finitely many intersections.
However, if you consider the problem in 2D complex projective space, then a beautiful theorem makes everything more harmonious: Bézout's theorem (algebraic geometry). According to this, if the set of intersection points is finite, then the number of intersections is simply equal to the product of the degrees of the varieties.
As a variety, the circle has degree 2. Therefore, for finitely many intersections, the number of intersections of any two circles, which may or may not be intersecting on the real plane, is 4. This may include real or complex points, coincident points (multiplicity), and even "points at infinity" (a specialty of projective spaces). In this manner, two lines will always meet at one point (1*1 = 1), a line and a circle at two (1*2 = 2), and so on. By the way, two ellipses can intersect at 4 real points, unlike circles which are only special ellipses.
A circle is an algebraic variety on the real plane. Two distinct circles will intersect at most at two points, as shown by other answers. However, this is not too satisfactory. Because, the circles sometimes meet at a single point, and sometimes never. Same is true for two lines: they may intersect at one point or they may not intersect at all. We are considering only finitely many intersections.
However, if you consider the problem in 2D complex projective space, then a beautiful theorem makes everything more harmonious: Bézout's theorem (algebraic geometry). According to this, if the set of intersection points is finite, then the number of intersections is simply equal to the product of the degrees of the varieties.
As a variety, the circle has degree 2. Therefore, for finitely many intersections, the number of intersections of any two circles, which may or may not be intersecting on the real plane, is 4. This may include real or complex points, coincident points (multiplicity), and even "points at infinity" (a specialty of projective spaces). In this manner, two lines will always meet at one point (1*1 = 1), a line and a circle at two (1*2 = 2), and so on. By the way, two ellipses can intersect at 4 real points, unlike circles which are only special ellipses.
Similar questions