Problem
GEO-B3-M01-P022 Apollonius Circle via Inversion
Given points \(A\) and \(B\) and a number \(k>0\), \(k\neq1\). Prove that the locus of points \(X\) such that \(\frac{XA}{XB}=k\) is a circle. Solve the problem using inversion centered at \(A\).
Hint 1. Choose the radius of inversion so that the image of \(B\) is convenient.
Hint 2. After inversion, the ratio \(XA:XB\) can be turned into a linear condition on \(X^*\).
E. Full solution. Take inversion centered at \(A\) with radius \(R\), where \(R^2=k\cdot AB^2\). Let \(X\) map to \(X^*\). From the similarity of inverse triangles we get \(\frac{XB}{XA}=\frac{X^*B^*}{AB^*}\), where \(B^*\) is the image of \(B\). The condition \(\frac{XA}{XB}=k\) is equivalent to \(\frac{XB}{XA}=\frac1k\), hence \(X^*B^*=\frac{AB^*}{k}\). Thus \(X^*\) lies on the circle centered at \(B^*\) with fixed radius \(\frac{AB^*}{k}\). This circle does not pass through center \(A\), so its inverse image is a circle. Therefore the original locus of points \(X\) is a circle.
This is stronger than the usual derivation of the Apollonius circle: the student sees how a distance ratio becomes a circle after a well-chosen inversion.