Power of a Point

Power of a Point — unifies "chord / secant / tangent through a fixed point" into a single product identity.

Definition of the power

For any point $P$ in the plane and a fixed circle $\odot O$ (radius $r$), define the power of $P$ with respect to $\odot O$ as

$\operatorname{pow}_{\odot O}(P) \;=\; |PO|^{2} - r^{2}.$

• $P$ outside the circle ⇒ $\operatorname{pow} > 0$
• $P$ on the circle ⇒ $\operatorname{pow} = 0$
• $P$ inside the circle ⇒ $\operatorname{pow} < 0$

The absolute value of the power has a clean geometric meaning — it is exactly the "product of the two segments cut on any chord / secant / tangent through $P$".

Three configurations (one theorem, three geometric guises)

Configuration 1 · Intersecting chords ($P$ inside)

Chord $AB$ and chord $CD$ meet at an interior point $P$:

$PA \cdot PB \;=\; PC \cdot PD \;=\; r^{2} - |PO|^{2}.$

Proof in Intersecting chords: PA·PB = PC·PD — draw auxiliary chords $AC, BD$; vertical angles + same-arc inscribed angles ⇒ $\triangle PAC \sim \triangle PDB$ ⇒ ratio ⇒ equal products.

Configuration 2 · Tangent + secant ($P$ outside)

From $P$ draw a tangent $PT$ (touching at $T$) and a secant $PAB$ (cutting at $A, B$):

$PT^{2} \;=\; PA \cdot PB \;=\; |PO|^{2} - r^{2}.$

Proof in Secant–tangent: tangent² = secant·external — tangent-chord angle + shared angle ⇒ $\triangle PTA \sim \triangle PBT$ ⇒ $PT/PB = PA/PT$.

Configuration 3 · Two secants ($P$ outside)

From $P$ draw two secants $PAB$ and $PCD$:

$PA \cdot PB \;=\; PC \cdot PD \;=\; |PO|^{2} - r^{2}.$

Proof in Two-secants product equality — same structure as the intersecting-chords version, with vertical angles replaced by a shared angle.

How to recognize / when to use

• The problem has two chords or secants through a common point and asks about a product of lengths or one unknown length → think Power of a Point
• The problem has a tangent + secant sharing an endpoint → use $PT^2 = PA \cdot PB$ directly
• Finding the extremum of a product on a moving chord through a fixed external point → the power is constant, independent of how the chord rotates
• Converse: given $PA \cdot PB = PC \cdot PD$ in the same configuration, the converse of this theorem proves the four points are concyclic.

Applications

To be added.

Related

• Intersecting chords: PA·PB = PC·PD / Secant–tangent: tangent² = secant·external / Two-secants product equality — rigorous proofs of the three configurations
• Tangent–chord angle = inscribed — the core lemma for the tangent-secant configuration
• AA similarity — the shared termination point of all three configurations' reasoning
• Miller's theorem — uses $CP^{*2} = CA \cdot CB$ to locate the tangent point of the max-angle circle