Prove that two different circles cannot intersect each other at more than two points.
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Proof by contradiction. Suppose to the contrary that two orthogonal circles did not meet at exactly 2 points. This leaves three cases. Case 0: The circles have zero points in common. These circles are disjoint and trivial. Case 1: The circles have one point in common. This by definition means that the circles are tangent to each other. Two circles that are tangent have the same tangent line at the point the circles are tangent. Thus the two circles can't orthogonal by definition. Case 3: The circles have three or more points in common. If two circles have at least 3 points in common then they are the same circle. These three points can't be collinear, since a line only intersects a circle twice. Since they are not collinear they form a triangle and both circles circumscribe the triangle. The circle that circumscribes any triangle is unique can be proved using the same ideas used in #11 regarding chords and their intersection with the circle's center. The results thus far have shown that two circles with zero points in common are disjoint, two circles with a single common point are tangent and two circles with three or more common points are the same circle, we must conclude that two orthogonal circles can only intersect in exactly two points.
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Two circles may intersect in two imaginary points, a single degenerate point, or two distinct points.
The intersections of two circles determine a line known as the radical line. If three circles mutually intersect in a single point, their point of intersection is the intersection of their pairwise radical lines, known as the radical center.
Let two circles of radii
Combining (1) and (2) gives
Multiplying through and rearranging gives
Solving for results in
The chord connecting the cusps of the lens therefore has half-length
Solving for
This same formulation applies directly to the sphere-sphere intersection problem.
To find the area of the asymmetric "lens" in which the circles intersect, simply use the formula for the circular segment of radius
twice, one for each half of the "lens." Noting that the heights of the two segment triangles are
The result is
The limiting cases of this expression can be checked to give 0 when
when
In order for half the area of two unit disks (
and solve numerically, yielding
If three symmetrically placed equal circles intersect in a single point, as illustrated above, the total area of the three lens-shaped regions formed by the pairwise intersection of circles is given by
Similarly, the total area of the four lens-shaped regions formed by the pairwise intersection of circles is given by
Borromean Rings, Brocard Triangles, Circle-Ellipse Intersection, Circle-Line Intersection, Circular Segment, Circular Triangle, Double Bubble, Goat Problem, Johnson's Theorem, Lens, Lune, Mohammed Sign, Moss's Egg, Radical Center, Radical Line, Reuleaux Triangle, Sphere-Sphere Intersection, Steiner Construction, Triangle Arcs, Triquetra, Venn Diagram, Vesica Piscis Sloane, N. J. A. Sequence A133741 in "The On-Line Encyclopedia of Integer Sequences."
Weisstein, Eric W. "Circle-Circle Intersection." From MathWorld--A Wolfram Web Resource. //mathworld.wolfram.com/Circle-CircleIntersection.html