Problem
GEO-B2-M02-P023 A Tangent From a Product
Point \(P\) lies outside a circle, and secant \(PAB\) meets it at \(A\) and \(B\), where \(A\) is closer to \(P\). Point \(T\) lies on the circle and satisfies \(PT^2=PA\cdot PB\). Prove that line \(PT\) is tangent to the circle.
If \(PT\) is not tangent, then it meets the circle at another point \(U\).
Assume that \(PT\) is not tangent. Then line \(PT\) meets the circle at another point \(U\). By the secant theorem, \(PT\cdot PU=PA\cdot PB\). By the condition, \(PT^2=PA\cdot PB\), hence \(PT\cdot PU=PT^2\), so \(PU=PT\). But on the ray from \(P\), the point of the circle at distance \(PT\) is already \(T\), so \(U=T\), contradicting two distinct intersections. Therefore \(PT\) is tangent to the circle.
The problem requires a careful converse understanding of the tangent formula.