Lemma › Techniques › Power of a Point
Olympiad problem-solving technique

Power of a Point

One number controls every line through P.

For a point PP and a circle of centre OO, radius rr, the power of PP is OP2−r2OP^2 - r^2. Every line through PP meeting the circle at AA and BB gives the same product PA⋅PBPA \cdot PB, and the choice of line is irrelevant, and that invariance is the technique.

It wears three faces: two secants (PA⋅PB=PC⋅PDPA \cdot PB = PC \cdot PD), a tangent and a secant (PT2=PA⋅PBPT^2 = PA \cdot PB), and two chords crossing inside. All three are the same statement.

The locus of equal power to two circles is a line, the radical axis, which is why power arguments so often deliver collinearity or concurrency without any computation.

Where it appears on Lemma: Level 5 (circles & power of a point) and Level 8 configurations.

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Power of a Point unlocks at Level 5 of Lemma's eight-level ladder, with lessons that teach it and drills that make it stick. 125 lessons, 1206 curated problems and unlimited generated practice at six difficulties. Free to start, no card, and every paid plan opens with 3 free days.

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