Problem
GEO-B2-M05-P021 Radius for an Invariant Circle
#21
★★★★★ Level 5 of 5
A circle \(\omega\) has centre \(C\), radius \(5\), and \(OC=13\). Find the radius of the inversion with centre \(O\) under which \(\omega\) maps to itself.
Use the power of point \(O\): \(R^2=OC^2-r^2\).
For the circle to be invariant, for every secant through \(O\) the product of distances to the intersection points must equal \(R^2\). This product is the power of the point: \(OC^2-r^2=13^2-5^2=169-25=144\). Hence \(R=12\).
A connection with the module on power of a point.