Problem
GEO-B3-M03-P001 The Basic Polar Formula
B. New Original Problem. From a point \(P\), tangents \(PA\) and \(PB\) are drawn to a circle \(\omega(O,R)\). Let \(H=AB\cap OP\). Prove that \(AB\perp OP\) and \(OP\cdot OH=R^2\).
C. Hint 1. Use congruence of the right triangles \(OAP\) and \(OBP\).
D. Hint 2. After proving perpendicularity, use the projection formula in a right triangle.
E. Full Solution.
The triangles \(OAP\) and \(OBP\) are right triangles, have the common hypotenuse \(OP\), and equal legs \(OA=OB=R\). Hence they are congruent, and \(A\) and \(B\) are symmetric with respect to \(OP\). Therefore \(AB\perp OP\).
In the right triangle \(OAP\), the point \(H\) is the foot of the altitude from \(A\) to the hypotenuse \(OP\). By the leg theorem, \(OA^2=OH\cdot OP\). Thus \(OP\cdot OH=R^2\).
This is the basic task: after it, the polar can be defined not only for an external point, but also by the formula.