#22 Prove if two circles are orthogonal, then they intersect at exactly two points and their tangents are perpendicular at both of those points
- Author:
- Bill OConnell
- Topic:
- Circle
#22 Prove if two circles are orthogonal, then they intersect at exactly two points and their tangents are perpendicular at both of those points
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.