# Simple Construction of the Circle of Apollonius

### Problem

### Solutions

$\mathcal{C}$ is the circle of Apollonius $\displaystyle \frac{XA}{XB}=\frac{CA}{CB}\,(\gt 1)$.

Indeed, with $\angle EDC=\angle BCA=2\varphi$ one has $\angle EFC=\angle BCE=\varphi$: $CE$ is the angle bisector of $\angle ACB$.

Another explanation is in that $A$ and $B$ are inverse images of each other about the circle of Apollonius $\mathcal{C}.$

### Acknowledgment

The above construction, with a proof, has been devised by Grégoire Nicollier (UNIVERSITY OF APPLIED SCIENCES OF WESTERN SWITZERLAND).

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