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Difference between revisions of "Power of a Point Theorem"

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The Power of a Point theorem expresses the relation between the lengths of two lines intersecting at a point and a [[circle]].  
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The '''Power of a Point Theorem''' is a relationship that holds between the lengths of the [[line segment]]s formed when two [[line]]s [[intersect]] a [[circle]] and each other.
  
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==See also==
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== Statement ==
 +
There are three possibilities as displayed in the figures below:
 +
# The two lines are [[secant line|chords]] of the circle and intersect inside the circle (figure on the left). In this case, we have <math> AE\cdot CE = BE\cdot DE </math>.
 +
# One of the lines is [[tangent line|tangent]] to the circle while the other is a [[secant line|secant]] (middle figure). In this case, we have <math> AB^2 = BC\cdot BD </math>.
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# Both lines are [[secant line|secants]] of the circle and intersect outside of it (figure on the right).  In this case, we have <math> CB\cdot CA = CD\cdot CE.</math>
 +
 
 +
[[Image:Pop.PNG|center]]
 +
 
 +
=== Hint for Proof===
 +
Draw extra lines to create similar triangles! (Hint: Draw <math>AD</math> on all three figures. Draw another line as well.)
 +
 
 +
=== Alternate Formulation ===
 +
This alternate formulation is much more compact, convenient, and general.
 +
 
 +
Consider a circle <math>O</math> and a point <math>P</math> in the plane where <math>P</math> is not on the circle. Now draw a line through <math>P</math> that intersects the circle in two places. The power of a point theorem says that the product of the length from <math>P</math> to the first point of intersection and the length from <math>P</math> to the second point of intersection is constant for any choice of a line through <math>P</math> that intersects the circle. This constant is called the power of point <math>P</math>. For example, in the figure below
 +
<cmath>
 +
PX^2=PA_1\cdot PB_1=PA_2\cdot PB_2=\cdots=PA_i\cdot PB_i
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</cmath>
 +
[[Image:Popalt.PNG|center]]
 +
 
 +
Notice how this definition still works if <math>A_k</math> and <math>B_k</math> coincide (as is the case with <math>X</math>). Consider also when <math>P</math> is inside the circle. The definition still holds in this case.
 +
 
 +
== Additional Notes ==
 +
One important result of this theorem is that both tangents from any point <math>P</math> outside of a circle to that circle are equal in length.
 +
 
 +
The theorem generalizes to higher dimensions, as follows.
 +
 
 +
Let <math>P</math> be a point, and let <math>S</math> be an <math>n</math>-sphere. Let two arbitrary lines passing through <math>P</math> intersect <math>S</math> at <math>A_1,B_1;A_2,B_2</math>, respectively. Then
 +
<cmath>PA_1\cdot PB_1=PA_2\cdot PB_2</cmath>
 +
 
 +
''Proof.'' We have already proven the theorem for a <math>1</math>-sphere (a circle), so it only remains to prove the theorem for more dimensions.  Consider the [[plane]] <math>p</math> containing both of the lines passing through <math>P</math>.  The intersection of <math>P</math> and <math>S</math> must be a circle.  If we consider the lines and <math>P</math> with respect simply to that circle, then we have reduced our claim to the case of two dimensions, in which we know the theorem holds.
 +
 
 +
== Problems ==
 +
 
 +
=== Introductory ===
 +
 
 +
* Find the value of <math>x</math> in the following diagram: [[Image:popprob1.PNG|center]]
 +
:[[Power of a Point Theorem/Introductory_Problem_1|Solution]]
 +
 
 +
* Find the value of <math>x</math> in the following diagram: [[Image:popprob2.PNG|center]]
 +
:[[Power of a Point Theorem/Introductory_Problem_2|Solution]]
 +
 
 +
* ([[ARML]]) In a circle, chords <math>AB</math> and <math>CD</math> intersect at <math>R</math>. If <math>AR:BR=1:4</math> and <math>CR:DR=4:9</math>, find the ratio <math>AB:CD</math>  .
 +
[[Image:popprob3.PNG|center]]
 +
:[[Power of a Point Theorem/Introductory_Problem_3|Solution]]
 +
 
 +
* ([[ARML]]) Chords <math>AB</math> and <math>CD</math> of a given circle are [[perpendicular]] to each other and intersect at a right angle at point <math>E</math>. Given that <math>BE=16</math>, <math>DE=4</math>, and <math>AD=5</math>, find <math>CE</math>.
 +
:[[Power of a Point Theorem/Introductory_Problem_4|Solution]]
 +
 
 +
=== Intermediate ===
 +
 
 +
* Two tangents from an external point <math>P</math> are drawn to a circle and intersect it at <math>A</math> and <math>B</math>.  A third tangent meets the circle at <math>T</math>, and the tangents <math>\overrightarrow{PA}</math> and <math>\overrightarrow{PB}</math> at points <math>Q</math> and <math>R</math>, respectively (this means that T is on the minor arc <math>AB</math>). If <math>AP = 20</math>, find the perimeter of <math>\triangle PQR</math>. ([[1961_AHSME_Problems/Problem_11|Source]])
 +
 
 +
* Square <math>ABCD</math> of side length <math>10</math> has a circle inscribed in it. Let <math>M</math> be the midpoint of <math>\overline{AB}</math>. Find the length of that portion of the segment <math>\overline{MC}</math> that lies outside of the circle. ([[2020 AMC 12B Problems/Problem 10|Source]])
 +
 
 +
* <math>DEB</math> is a chord of a circle such that <math>DE=3</math> and <math>EB=5 .</math> Let <math>O</math> be the center of the circle. Join <math>OE</math> and extend <math>OE</math> to cut the circle at <math>C.</math> Given <math>EC=1,</math> find the radius of the circle. ([[1971_Canadian_MO_Problems/Problem_1|Source]])
 +
[[Image:CanadianMO_1971-1.jpg]]
 +
 
 +
* Triangle <math>ABC</math> has <math>BC=20.</math> The incircle of the triangle evenly trisects the median <math>AD.</math> If the area of the triangle is <math>m \sqrt{n}</math> where <math>m</math> and <math>n</math> are integers and <math>n</math> is not divisible by the square of a prime, find <math>m+n.</math> ([[2005 AIME I Problems/Problem 15|Source]])
 +
 
 +
=== Olympiad ===
 +
 
 +
* {{incomplete}}
 +
 
 +
== See Also ==
 
* [[Geometry]]
 
* [[Geometry]]
 
* [[Planar figures]]
 
* [[Planar figures]]
 +
* [{{SERVER}}/community/c6h2513977 Kagebaka's Handout]
 +
 +
[[Category:Geometry]]
 +
[[Category:Theorems]]
 
{{stub}}
 
{{stub}}

Latest revision as of 09:36, 16 February 2025

The Power of a Point Theorem is a relationship that holds between the lengths of the line segments formed when two lines intersect a circle and each other.

Statement

There are three possibilities as displayed in the figures below:

  1. The two lines are chords of the circle and intersect inside the circle (figure on the left). In this case, we have $AE\cdot CE = BE\cdot DE$.
  2. One of the lines is tangent to the circle while the other is a secant (middle figure). In this case, we have $AB^2 = BC\cdot BD$.
  3. Both lines are secants of the circle and intersect outside of it (figure on the right). In this case, we have $CB\cdot CA = CD\cdot CE.$
Pop.PNG

Hint for Proof

Draw extra lines to create similar triangles! (Hint: Draw $AD$ on all three figures. Draw another line as well.)

Alternate Formulation

This alternate formulation is much more compact, convenient, and general.

Consider a circle $O$ and a point $P$ in the plane where $P$ is not on the circle. Now draw a line through $P$ that intersects the circle in two places. The power of a point theorem says that the product of the length from $P$ to the first point of intersection and the length from $P$ to the second point of intersection is constant for any choice of a line through $P$ that intersects the circle. This constant is called the power of point $P$. For example, in the figure below \[PX^2=PA_1\cdot PB_1=PA_2\cdot PB_2=\cdots=PA_i\cdot PB_i\]

Popalt.PNG

Notice how this definition still works if $A_k$ and $B_k$ coincide (as is the case with $X$). Consider also when $P$ is inside the circle. The definition still holds in this case.

Additional Notes

One important result of this theorem is that both tangents from any point $P$ outside of a circle to that circle are equal in length.

The theorem generalizes to higher dimensions, as follows.

Let $P$ be a point, and let $S$ be an $n$-sphere. Let two arbitrary lines passing through $P$ intersect $S$ at $A_1,B_1;A_2,B_2$, respectively. Then \[PA_1\cdot PB_1=PA_2\cdot PB_2\]

Proof. We have already proven the theorem for a $1$-sphere (a circle), so it only remains to prove the theorem for more dimensions. Consider the plane $p$ containing both of the lines passing through $P$. The intersection of $P$ and $S$ must be a circle. If we consider the lines and $P$ with respect simply to that circle, then we have reduced our claim to the case of two dimensions, in which we know the theorem holds.

Problems

Introductory

  • Find the value of $x$ in the following diagram:
    Popprob1.PNG
Solution
  • Find the value of $x$ in the following diagram:
    Popprob2.PNG
Solution
  • (ARML) In a circle, chords $AB$ and $CD$ intersect at $R$. If $AR:BR=1:4$ and $CR:DR=4:9$, find the ratio $AB:CD$ .
Popprob3.PNG
Solution
  • (ARML) Chords $AB$ and $CD$ of a given circle are perpendicular to each other and intersect at a right angle at point $E$. Given that $BE=16$, $DE=4$, and $AD=5$, find $CE$.
Solution

Intermediate

  • Two tangents from an external point $P$ are drawn to a circle and intersect it at $A$ and $B$. A third tangent meets the circle at $T$, and the tangents $\overrightarrow{PA}$ and $\overrightarrow{PB}$ at points $Q$ and $R$, respectively (this means that T is on the minor arc $AB$). If $AP = 20$, find the perimeter of $\triangle PQR$. (Source)
  • Square $ABCD$ of side length $10$ has a circle inscribed in it. Let $M$ be the midpoint of $\overline{AB}$. Find the length of that portion of the segment $\overline{MC}$ that lies outside of the circle. (Source)
  • $DEB$ is a chord of a circle such that $DE=3$ and $EB=5 .$ Let $O$ be the center of the circle. Join $OE$ and extend $OE$ to cut the circle at $C.$ Given $EC=1,$ find the radius of the circle. (Source)

CanadianMO 1971-1.jpg

  • Triangle $ABC$ has $BC=20.$ The incircle of the triangle evenly trisects the median $AD.$ If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n.$ (Source)

Olympiad

See Also

This article is a stub. Help us out by expanding it.

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