Line Through a Center of Similarity

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28 June 2014, Created with GeoGebra

Problem

$P$ is a center of similarity of circles $(O_{1})$ and $(O_{2}),$ centered at $O_{1}$ and $O_{2},$ respectively. A line through $P$ meets $(O_{i})$ at points $A_{i}$ and $B_{i},$ $i=1,2.$ The tangents to the circles, one at $A_{1},$ the other at $B_{2},$ meet at $X,$ those at $A_{2}$ and $B_{1}$ meet at $Y.$

Line Through a Center of Similarity - problem

Prove that $XY$ is the radical axis of the two circles.

Solution

Since the given line passes through a center of similarity of the two circles, $A_{1},$ say, maps on $A_{2}$ and $B_{1}$ on $B_{2}.$ Naturally, $O_{1}$ maps on $O_{2}.$

Line Through a Center of Similarity - solution

This, in particular, means that $\angle A_{1}O_{1}B_{1}=\angle A_{2}O_{2}B_{2}$ and that the pairs of the tangents are parallel: $A_{1}X\parallel A_{2}Y$ and $B_{2}X\parallel B_{1}Y.$ The angle between a chord and a tangent at one of the chord's endpoints equals half the central angle subtended by the chord. Thus, $\angle A_{1}B_{2}X=\angle B_{2}A_{1}X,$ implying $A_{1}X=B_{2}X.$ Similarly, $A_{2}Y=B_{1}Y,$ which exactly means that both $X$ and $Y$ lie on the radical axis of the to circles.

The diagram above illustrates the case of the external center of similarity. The case of the internal center is entirely similar.

Note: The case where the given line does not pass through one of the centers of similarity is treated on a separate page.

Acknowledgment

This is problem 251a from the second volume of I. M. Yaglom's Geometric Transformations, Russian edition (1956.) I am pretty sure that the problem has been included in the fourth volume of the English translation, but not owning the book I am unable to verify that fact.

Radical Axis and Radical Center

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