Coaxal Circles Theorem

The family of the Apollonian circles defined by two points, say, A and B have a few characteristic properties:

1. No two of the circles intersect.

2. Centers of all the circles in the family are collinear.

3. All the circles are orthogonal to the common line of centers.

4. None of the circles has its center between A and B.

5. The family of circles contains one straight line: the perpendicular bisector of AB.

6. Any two of the circles have the same radical axis: the straight line member of the family. (This will be proven shortly.)

1. Families in which the circles have no common points (these are the Apollonian circles of a pair of distinct points.)

2. Families that share exactly one point.

3. Families that share exactly two points.

The Coaxal Circles Theorem asserts that the families come out in pairs: all circles in one family is orthogonal to all the circles in the other. The pairing is simple.

Coaxal Circles Theorem

The second part is obvious as of the two perpendicular lines (axes of both families), one is tangent to all the circles in one family, the other is tangent to all the circles in the other family.

To prove the first part, consider the inversion t in the unit circle centered at, say, A, the way we did in the discussion on the Apollonian circles. Any circle through A inverts into a straight line. Any circle through A and B inverts into a straight line through t(B). On the other hand, the Apollonian circles invert into the family of circles centered at t(B). These circles and the lines through t(B) are obviously orthogonal. Since, an inversion, as is its inverse, is an angle preserving transformation the original families of circles are also orthogonal. (The applet showing the Apollonian circles provides a clear visualization of this property. Just choose N = 10 or more and rotate P.)

(The families of circles paired via the Coaxal Circles Theorem are known as conjugate.)

The Coaxal Circles Theorem yields an easy proof of the #6 property of the Apollonian circles. Pick a point on the perpendicular bisector of AB and consider circle C from the conjugate family centered at this point. C is orthogonal to any circle from the given Apollonian family. This in particular means that the radius-vector of any of the circles to the point common with C is orthogonal to the corresponding radius-vector of C to the same point. It follows that the latter is tangent to the circle at hand and that the square of its radius is exactly the power of its center with respect to any of the circles in the Apollonian family.

References

1. D. Pedoe, Circles: A Mathematical View, MAA, 1995

Radical Axis and Radical Center

• A Property of the Line IO: A Proof From The Book
• Cherchez le quadrilatère cyclique II
• Circles On Cevians
• Circles And Parallels
• Circles through the Orthocenter
• Coaxal Circles Theorem
• Isosceles on the Sides of a Triangle
• Properties of the Circle of Similitude
• Six Concyclic Points
• Radical Axis and Center, an Application
• Radical axis of two circles
• Radical Axis of Circles Inscribed in a Circular Segment
• Radical Center
• Radical center of three circles
• Steiner's porism
• Stereographic Projection and Inversion
• Stereographic Projection and Radical Axes
• Tangent as a Radical Axis
• Two Circles on a Side of a Triangle
• Pinning Butterfly on Radical Axes

Copyright © 1996-2009 Alexander Bogomolny