# Power of a Point Theorem

## Theorem

Let $C$ be a circle in the Euclidean plane whose center is $O$ and whose radius is $r$.

Let $P$ be an arbitrary point in the plane.

Let $p$ be the power of point $P$ with respect to $C$.

Let a directed line segment from $P$ be drawn either:

intersecting $C$ at two points $A$ and $A'$

or:

tangent to $C$ at $A = A'$.

Then:

$PA \cdot PA' = p$

## Proof

Let $d$ be the distance from $P$ to $O$.

Let $t$ be the length of the tangent from $P$ to $C$.

We use the following several times.

 $\ds PA$ $=$ $\ds -AP$ Definition of Directed Line Segment $\text {(1)}: \quad$ $\ds PA \cdot PA'$ $=$ $\ds -AP \cdot PA'$ substitution $\text {(2)}: \quad$ $\ds p$ $=$ $\ds d^2 - r^2$ Definition of Power of Point $\text {(3)}: \quad$ $\ds d^2$ $=$ $\ds t^2 + r^2$ Pythagoras's Theorem $\text {(4)}: \quad$ $\ds p$ $=$ $\ds t^2$ from $(2)$ and $(3)$

$P$ is Circle Center

By definition of center of circle:

a straight line through $O$ is a diameter of $C$.
 $\ds P$ $=$ $\ds O$ by construction $\ds \leadsto \ \$ $\ds d$ $=$ $\ds 0$ $\ds AP$ $=$ $\ds r$ Definition of Circle $\ds PA'$ $=$ $\ds r$ Definition of Circle $\ds \leadsto \ \$ $\ds AP \cdot PA'$ $=$ $\ds r^2$ $\ds PA \cdot PA'$ $=$ $\ds -r^2$ $(1)$ $\ds p$ $=$ $\ds d^2 - r^2$ $(2)$ $\ds \leadsto \ \$ $\ds p$ $=$ $\ds -r^2$ $\ds \leadsto \ \$ $\ds PA \cdot PA'$ $=$ $\ds p$

$\Box$

$P$ is Interior Point of $C$

Let $P$ be in the interior of $C$ but not at $O$.

Let $BB'$ be the diameter of $C$ through $P$.

 $\ds BP$ $=$ $\ds r - d$ by construction $\ds PB'$ $=$ $\ds r + d$ by construction $\ds \leadsto \ \$ $\ds BP \cdot PB'$ $=$ $\ds \paren {r + d} \paren {r - d}$ $\ds \leadsto \ \$ $\ds BP \cdot PB'$ $=$ $\ds r^2 - d^2$ $\ds AP \cdot PA'$ $=$ $\ds BP \cdot PB'$ Intersecting Chords Theorem $\ds AP \cdot PA'$ $=$ $\ds r^2 - d^2$ Common Notion $1$ $\ds PA \cdot PA'$ $=$ $\ds d^2 - r^2$ $(1)$ $\ds p$ $=$ $\ds d^2 - r^2$ $(2)$ $\ds \leadsto \ \$ $\ds PA \cdot PA'$ $=$ $\ds p$

$\Box$

$P$ is on circle $C$
 $\ds PA$ $=$ $\ds 0$ by construction $\ds \leadsto \ \$ $\ds PA \cdot PA'$ $=$ $\ds 0$ $\ds d$ $=$ $\ds r$ by construction $\ds p$ $=$ $\ds d^2 - r^2$ $(2)$ $\ds \leadsto \ \$ $\ds p$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds PA \cdot PA'$ $=$ $\ds p$

$\Box$

$P$ is Exterior Point and $PA$ tangent to $C$

Let $A$ be such that $PA$ is tangent to $C$ at $A$.

We have by hypothesis that the point $A'$ is the same point as $A$.

 $\ds PA$ $=$ $\ds t$ by construction $\ds PA'$ $=$ $\ds PA$ by hypothesis $\ds \leadsto \ \$ $\ds PA \cdot PA'$ $=$ $\ds t^2$ $\ds p$ $=$ $\ds t^2$ $(4)$ $\ds \leadsto \ \$ $\ds PA \cdot PA'$ $=$ $\ds p$

$\Box$

$P$ is Exterior Point and $PAA'$ a secant of $C$
 $\ds PA \cdot PA'$ $=$ $\ds t^2$ Tangent Secant Theorem $\ds p$ $=$ $\ds t^2$ $(4)$ $\ds \leadsto \ \$ $\ds PA \cdot PA'$ $=$ $\ds p$

The result follows.

$\blacksquare$