The Geometry Junkyard

Tangencies: Circular Angle Bisectors

Given any two crossing circles A and B (blue), there exist two more circles C and D (red) through the two crossing points, that bisect the angles made by A and B at those points. If any two circles E and F (green) are tangent to each other, with E and F both tangent to both of A and B, then the tangency between E and F lies on one of C or D. Further, if (say) the tangency lies on C, then C is perpendicular to both E and F.

Circular angle bisectors
If you were running Java, you'd see a nice animation instead of this gif.

As usual, this can be proven easily by inversion: just invert by a circle centered on one of the crossing points of A and B, so that A and B are transformed to two crossing lines, and C and D are transformed to the two angle bisectors of those lines.

Note that the angle bisectors, and their property of containing the tangents of pairs of disks tangent to each other and to A and B, continue to exist even when A and B do not cross, even though in this case there is no longer any angle to bisect. In this case, the existence of these circles is again seen by inversion, to a configuration in which A and B are concentric. The existence of these bisectors is especially useful for constructing instances of Steiner's porism, rings of tangent circles between two circles, since it gives a point on each successive tangent circle from the previous one.

If we are given any point x on C (or D), we can construct the tangent circles E and F using the four-circle property: the four circles A, E, the line L through x and the center of C (viewed as an infinite-radius circle), and a circle along the line perpendicular to L through the center of A together form a cycle of four tangent circles, three tangencies of which are known or easily constructed. The fourth tangency, of A with E, is then the point where the circle through these three tangencies crosses A. Once we have similarly constructed the tangency of B with E, we have three points on E from which the circle itself is easily found.

Animation created by Cinderella.
From the Geometry Junkyard, computational and recreational geometry.
David Eppstein, Theory Group, ICS, UC Irvine.

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