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 $C$
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
- Intersecting Chords Theorem, of which the Power of a Point Theorem is a generalization
Sources
- Weisstein, Eric W. "Circle Power." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/CirclePower.html