Problem
GEO-B2-M05-P017 Formula for a Circle Through the Centre
#17
★★★★☆ Level 4 of 5
A circle \(\omega\) passes through \(O\), has centre \(C\), and radius \(r\). The inversion has radius \(R\). Prove that the image of \(\omega\) is a line perpendicular to \(OC\), at distance \(\frac{R^2}{2r}\) from \(O\).
Take the point \(Q\) of the circle on ray \(OC\), different from \(O\). Then \(OQ=2r\).
Point \(Q\) maps to \(Q'\), where \(OQ\cdot OQ'=R^2\). Since \(OQ=2r\), we have \(OQ'=\frac{R^2}{2r}\). A circle through \(O\) maps to a line; in this configuration, that line passes through \(Q'\) perpendicular to \(OC\). Therefore the distance from \(O\) to the image line is \(\frac{R^2}{2r}\).
The formula is useful for quick computations.