Problem
GEO-B2-M02-P011 A Tangent and a Diameter Secant
#11
★★★☆☆ Level 3 of 5
From point \(P\) outside a circle, tangent \(PT=15\) is drawn. A secant through the centre of the circle meets the circle at \(A\) and \(B\), where \(A\) is closer to \(P\), and \(PA=9\). Find the diameter of the circle.
First find \(PB\), then \(AB=PB-PA\).
By the tangent-secant formula, \(PT^2=PA\cdot PB\). Thus \(225=9\cdot PB\), so \(PB=25\). The secant passes through the centre, hence \(AB\) is a diameter. We get \(AB=PB-PA=25-9=16\).
The problem connects power of a point with the geometric meaning of a secant through the centre.