The Radical Axis of Circles Inscribed in a Circular Segment: What is this about?
A Mathematical Droodle

Explanation

Copyright © 1996-2009 Alexander Bogomolny

Circles C and C' are homothetic with center B, such that A, B, and M are collinear. Triangles BTM and MAT are similar. Indeed, the two triangles share an angle at M. Also, BTM equals half the measure of the arc BSM. On the other hand, MAT equals half the measure of the sum of arcs BS and MT, but the latter is equal to MS. Therefore, BTM = MAT.

From the similarity of triangles BTM and MAT we obtain MB/MT = MT/MA, which is (2).

Now for the proof of (1). Observe that (2) holds for both C' and C'', as it does for any circle inscribed into the given circular segment. This means that M has the same power with respect to C' as it has with respect to C''. The points with that property lie on the radical axis of the two circles. The radical axis of two intersecting circles (that contains M) is the line that passes through the two points of intersection of C' and C''.

(The problem also has a simple interpretation in terms of the inversion transformation. See Problem 3 in the Inversion page.)

Inversion - Introduction

• Angle Preservation Property
• Apollonian Circles Theorem
• Archimedes' Twin Circles and a Brother
• Bisectal Circle
• Chain of Inscribed Circles
• Circle Inscribed in a Circular Segment
• Circle Inversion: Reflection in a Circle
• Circle Inversion Tool
• Feuerbach's Theorem: a Proof
• Four Touching Circles
• Hart's Inversor
• Inversion in the Incircle
• Inversion with a Negative Power
• Miquel's Theorem for Circles
• Peaucellier Linkage
• Polar Circle
• Poles and Polars
• Radical Axis of Circles Inscribed in a Circular Segment
• Steiner's porism
• Stereographic Projection and Inversion
• Tangent Circles and an Isosceles Triangle
• Tangent Circles and an Isosceles Triangle II
• Three Tangents, Three Secants

Copyright © 1996-2009 Alexander Bogomolny