# Line Through a Center of Similarity

### What is this about?

### 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 $Y.$

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 $O_{2}.$

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.

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