Problem
GEO-B3-M03-P021 Polar of an Interior Point via a Projective Model
B. New Original Problem. A point \(P\) lies inside a circle \(\omega(O,R)\), with \(P\ne O\). A line \(p\) is perpendicular to \(OP\) and meets \(OP\) at \(H\), where \(OP\cdot OH=R^2\). Prove that for every chord \(AB\) through \(P\), the intersection of the tangents at \(A\) and \(B\) lies on \(p\).
C. Hint 1. Even for an interior point, the polar is given by the same formula.
D. Hint 2. Apply La Hire to the tangent intersection and the secant \(AB\).
E. Full Solution.
By the formula \(OP\cdot OH=R^2\) and perpendicularity \(p\perp OP\), the line \(p\) is the polar of \(P\), even though no real tangents can be drawn from \(P\).
Let a chord \(AB\) pass through \(P\), and let the tangents at \(A\) and \(B\) meet at \(T\). The polar of \(T\) is \(AB\). Since \(P\in AB\), La Hire's theorem gives that \(T\) lies on the polar of \(P\), namely on \(p\). This holds for every chord \(AB\) through \(P\).
The task is important because it removes the false association of polars only with exterior points.