Power of a Point Theorem

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 unique cases for this theorem. Each case expresses the relationship between the length of line segments that pass through a common point and touch a circle in at least one point. Can be useful with cyclic quadrilaterals as well however with a slightly different application.

Case 1 (Inside the Circle):

If two chords $AB$ and $CD$ intersect at a point $P$ within a circle, then $AP\cdot BP=CP\cdot DP$

$[asy] draw(circle((0,0),3)); dot((-2.82,1)); label("A",(-3.05,1.25)); dot((1,2.828)); label("B",(1.25,3.05)); draw((-2.82,1)---(1,2.828)); dot((2.3,-1.926)); label("C",(2.55,-2.346)); dot((-2.12,2.123)); label("D",(-2.37,2.507)); draw((2.3,-1.926)---(-2.12,2.123)); dot((-1.556,1.602)); label("P",(-1.656,1.202)); [/asy]$

Case 2 (Outside the Circle):

Classic Configuration

Given lines $BP$ and $CP$ originate from two unique points on the circumference of a circle ( $B$ and $C$ ), intersect each other at point $P$ , outside the circle, and re-intersect the circle at points $A$ and $D$ respectively, then $PA\cdot PB=PD\cdot PC$

$[asy] draw(circle((0,0),3)); dot((1.5,2.598)); label("B",(2,3)); label("P",(-6,1.6)); dot((-6,1)); label("C",(2.55,-2.5)); dot((2.12,-2.123)); dot((-2.996,-0.155)); label("D",(-3.350, -0.6)); dot((-2.429,1.761)); label("A",(-2.729,2.061)); draw((1.5,2.598)---(-6,1)); draw((2.12,-2.123)---(-6,1)); [/asy]$

Tangent Line

Given Lines $AB$ and $AC$ with $AC$ tangent to the related circle at $C$ , $A$ lies outside the circle, and Line $AB$ intersects the circle between $A$ and $B$ at $D$ , $AD\cdot AB=AC^{2}$

$[asy] draw(circle((0,0),3)); dot((0,3)); label("C",(0,3.5)); dot((-8,3)); label("A",(-8,3.5)); dot((2.5,-1.658)); label("B",(2.8,-1.958)); draw((0,3)---(-8,3)); draw((2.5,-1.658)---(-8,3)); dot((-2.907,0.741)); label("D",(-3.357,0.421)); [/asy]$

Case 3 (On the Border/Useless Case):

If two chords, $AB$ and $AC$ , have $A$ on the border of the circle, then the same property such that if two lines that intersect and touch a circle, then the product of each of the lines segments is the same. However since the intersection points lies on the border of the circle, one segment of each line is $0$ so no matter what, the constant product is $0$ .

$[asy] draw(circle((0,0),3)); dot((1,2.828)); label("A",(1.4,3.028)); dot((-2.5,-1.658)); label("B",(-2.8,-1.958)); dot((2.04,-2.2)); label("C",(2.34,-2.5)); draw((1,2.828)---(-2.5,-1.658)); draw((1,2.828)---(2.04,-2.2)); [/asy]$

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 $PX^2=PA_1\cdot PB_1=PA_2\cdot PB_2=\cdots=PA_i\cdot PB_i$

Hint for Proof

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

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.

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.

Proof

Case 1 (Inside the Circle)

Join $AD$ and $BC$ .

In $\triangle ADP \; \text{and} \; \triangle CBP$

$\angle ADC = \angle CBA \hspace{1cm}$ (Angles subtended by the same segment are equal)

$\angle DPA = \angle BPC \hspace{1cm}$ (Vertically opposite angles)

$\therefore \; \triangle ADP \sim \triangle CBP$

$\implies \frac{AP}{CP} = \frac{DP}{BP} \hspace{1cm}$ (Corresponding sides of similar triangles are in the same ratio)

$\implies AP \cdot BP = DP \cdot CP$

$\blacksquare$

Case 2 (Outside the Circle)

Join $AD$ and $BC$

$\angle DAB + \angle DCB = 180^{\circ} = \angle PAD + \angle DAB \hspace{1cm}$ (Why?)

$\implies \angle PCB = \angle DCB = \angle PAD$

Now, In $\triangle PAD \; \text{and} \; \triangle PCB$

$\angle PAD = \angle PCB \hspace{1cm}$ (shown above)

$\angle APD = \angle CPB \hspace{1cm}$ (common angle)

$\therefore \; \triangle PAD \sim \triangle PCB$

$\implies \frac{PA}{PC} = \frac{PD}{PB} \hspace{1cm}$ (Corresponding sides of similar triangles are in the same ratio)

$\implies PA \cdot PB = PD \cdot PC$

$\blacksquare$

Case 3 (On the Circle Border)

Length of a point is zero so no proof needed :)

Problems

Introductory

Find the value of $x$ in the following diagram:

Solution

Find the value of $x$ in the following diagram:

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 $AB:CD$ .

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

Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$ (Source)

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)

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)

Let $ABC$ be a triangle inscribed in circle $\omega$ . Let the tangents to $\omega$ at $B$ and $C$ intersect at point $D$ , and let $\overline{AD}$ intersect $\omega$ at $P$ . If $AB=5$ , $BC=9$ , and $AC=10$ , $AP$ can be written as the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime integers. Find $m + n$ . (Source)

Olympiad

Given circles $\omega_1$ and $\omega_2$ intersecting at points $X$ and $Y$ , let $\ell_1$ be a line through the center of $\omega_1$ intersecting $\omega_2$ at points $P$ and $Q$ and let $\ell_2$ be a line through the center of $\omega_2$ intersecting $\omega_1$ at points $R$ and $S$ . Prove that if $P, Q, R$ and $S$ lie on a circle then the center of this circle lies on line $XY$ .

(Source)

Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$ ). The lines $AP$ , $BP$ and $CP$ meet again its circumcircle $\Gamma$ at $K$ , $L$ , respectively $M$ . The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$ . Show that from $SC = SP$ follows $MK = ML$ .