Problem
GEO-B2-M03-P006 Equal Products
#6
★★★☆☆ Level 3 of 5
From point \(P\), secant \(PAB\) is drawn to circle \(\omega_1\), and secant \(PCD\) to circle \(\omega_2\). It is known that \(PA\cdot PB=PC\cdot PD\). Prove that \(P\) lies on the radical axis of \(\omega_1\) and \(\omega_2\).
Each product is the power of point \(P\) with respect to the corresponding circle.
The product \(PA\cdot PB\) is the power of point \(P\) with respect to \(\omega_1\). The product \(PC\cdot PD\) is the power of point \(P\) with respect to \(\omega_2\). By the condition these powers are equal, so \(P\) lies on the radical axis of the two circles.
The problem trains recognition of a hidden radical axis without a common chord.