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Copyright © 1996-2008 Alexander Bogomolny
The Radical Axis of Circles Inscribed in a Circular SegmentThe applet suggests the following statement:
In order to prove (1), let's establish the following identity
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, From the similarity of triangles BTM and MAT we obtain 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
Copyright © 1996-2008 Alexander Bogomolny
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