Problem
GEO-B2-M03-P010 Any Secants From a Point on the Radical Axis
#10
★★★☆☆ Level 3 of 5
Point \(P\) lies on the radical axis of circles \(\omega_1\) and \(\omega_2\). Secants \(PAB\) to \(\omega_1\) and \(PCD\) to \(\omega_2\) are drawn through \(P\). Prove that \(PA\cdot PB=PC\cdot PD\).
On the radical axis, the powers of the point with respect to the two circles are equal.
Since \(P\) lies on the radical axis, \(\operatorname{Pow}_{\omega_1}(P)=\operatorname{Pow}_{\omega_2}(P)\). But \(\operatorname{Pow}_{\omega_1}(P)=PA\cdot PB\), and \(\operatorname{Pow}_{\omega_2}(P)=PC\cdot PD\). Therefore \(PA\cdot PB=PC\cdot PD\).
The reverse of the previous problem: first the radical axis, then products.