Power of a Point Theorem

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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 $C$
Power of Point Interior.png

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.

$\blacksquare$


Sources


Sources

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