Chapter 17 — Power of a Point
In this chapter
- 17.1 Introduction — The Theorem & Four Configurations
- 17.2 Power of a Point Proofs
- Key Formulas Table
- Exercises & Problems
Key Formulas
| Configuration | Diagram Description | Formula |
|---|---|---|
| Two tangents from point $A$ | Tangent to circle at $B$ and $C$ | $AB = AC$ |
| Tangent + secant from point $A$ | Tangent at $C$, secant through $B$ to $D$ | $AC^2 = (AB)(AD)$ |
| Two secants from point $A$ | Secant through $B,C$ and secant through $D,E$ | $(AB)(AC) = (AD)(AE)$ |
| Intersecting chords at point $A$ | Chords $BC$ and $DE$ crossing at $A$ | $(BA)(AC) = (DA)(AE)$ |
17.1 Introduction
Power of a Point Theorem. Given a point $P$ and a line through $P$ which intersects a circle at two points $A$ and $B$, the product $(PA)(PB)$ is the same for any choice of line through $P$.
This constant product is called the power of the point $P$ with respect to the circle.
The theorem has four major configurations:
Configuration 1: Two Tangents from a Point
Two tangents from the same external point to a circle are always equal:
\(AB = AC\)
Configuration 2: Tangent and Secant from a Point
Given tangent $AC$ (tangent at $C$) and secant through $A$ hitting the circle at $B$ then $D$:
\(AC^2 = (AB)(AD)\)
Configuration 3: Two Secants from a Point
Given secants from external point $A$, one through $B$ and $C$ on the circle, the other through $D$ and $E$:
\((AB)(AC) = (AD)(AE)\)
Configuration 4: Intersecting Chords
Given two chords $BC$ and $DE$ which intersect at interior point $A$:
\((BA)(AC) = (DA)(AE)\)
When to use Power of a Point:
- You see intersecting chords in a circle.
- You see a tangent and a secant from the same point.
- You see two tangents from one point — immediately mark them equal.
- You need ratios of segments involving a circle.
Power of a Point is generally easy to spot — if you have chords, tangents, or secants involving a circle, check for it!
Examples
Example 17-1. Given tangent $AC$ and secant $AB$ with $AC = 6$, $AD = 4$, $BD = x$. Find $x$.
Solution. By Power of a Point (tangent-secant): $(AB)(AD) = AC^2$. Here $AB = AD + DB = 4 + x$, so:
\(4(4 + x) = 36 \implies 16 + 4x = 36 \implies x = \boxed{5}\)
Example 17-2. Two diagonals $AX$ and $BY$ of a regular polygon intersect at $W$. Prove $(AW)(WX) = (BW)(WY)$.
Proof. Since the polygon is regular, all its vertices lie on a single circle (the circumscribed circle). Diagonals $AX$ and $BY$ are chords of this circle intersecting at $W$. By Power of a Point (intersecting chords):
\((AW)(WX) = (BW)(WY) \qquad \square\)
Example 17-3. In $\triangle ABC$, points $X$, $Y$, $Z$ are where the incircle is tangent to the sides ($X$ opposite $A$, $Y$ opposite $B$, $Z$ opposite $C$). Prove $AZ = s - a$, $BX = s - b$, $CY = s - c$, where $s$ is the semi-perimeter.
Proof. Since tangents from a point to a circle are equal:
- Let $AZ = AY = x$
- Let $BZ = BX = y$
- Let $CY = CX = z$
The perimeter is:
\[2s = (x + y) + (y + z) + (x + z) = 2(x + y + z)\]so $s = x + y + z$. Since $x + z = b$ (side $AC$), we get $y = BX = s - b$. Similarly $x = AZ = s - a$ and $z = CY = s - c$. $\square$
Example 17-4. Prove that if quadrilateral $ABCD$ can be circumscribed about a circle, then $AB + CD = BC + AD$.
Proof. Since the quadrilateral is circumscribed about a circle, each side is tangent to the circle. Label the tangent lengths from each vertex:
- From $A$: segments of length $w$ and $x$
- From $B$: segments of length $x$ and $y$
- From $C$: segments of length $y$ and $z$
- From $D$: segments of length $z$ and $w$
(using the fact that tangents from a point are equal). Then:
\(AB + CD = (x + w) + (y + z) = (w + z) + (x + y) = AD + BC \qquad \square\)
Exercises
Exercise 17-1. Prove that the inradius of a right triangle with legs $a$, $b$ and hypotenuse $c$ is $r = \frac{a + b - c}{2}$.
Exercise 17-2. Show that if $AB$ is a diameter of a circle and $CD$ is a chord perpendicular to $AB$ intersecting at $X$, then $CX^2 = (AX)(BX)$.
Converse use. Power of a Point can also be used to prove that a segment is tangent to a circle. If a line through point $A$ outside circle $O$ meets the circle at $B$ and $C$, and another line through $A$ meets the circle at $X$ such that $AX^2 = (AB)(AC)$, then $AX$ is tangent to the circle.
17.2 Power of a Point Proofs
The proofs use similar triangles (since the formulas involve ratios of sides) and arc-angle relationships (since circles are involved).
Case 1: Point Outside the Circle (Secant–Tangent)
Goal. Show $(AD)(AC) = AB^2$ where $AB$ is tangent and $ADC$ is a secant.
Rearranging: $\frac{AD}{AB} = \frac{AB}{AC}$, so we need $\triangle ADB \sim \triangle ABC$.
- $\angle DAB = \angle CAB$ (same angle).
- $\angle BCD = \frac{\widehat{BD}}{2}$ (inscribed angle).
- $\angle ABD = \frac{\widehat{BD}}{2}$ (tangent-chord angle).
So $\angle BCD = \angle ABD$, giving $\triangle ADB \sim \triangle ABC$ by AA. Therefore:
\[\frac{AD}{AB} = \frac{AB}{AC} \implies (AD)(AC) = AB^2\]Since this holds for any secant, $(AD)(AC)$ is constant for all secants through $A$. For two tangents $AB$ and $AE$: $AB^2 = (AD)(AC) = AE^2$, so $AB = AE$. $\square$
Case 2: Point Inside the Circle (Intersecting Chords)
Goal. Show $(AE)(BE) = (CE)(DE)$ where chords $AB$ and $CD$ intersect at $E$.
- $\angle AEC = \angle DEB$ (vertical angles).
- $\angle CAB = \angle CDB$ (inscribed angles subtending the same arc $BC$).
So $\triangle EAC \sim \triangle EDB$ by AA. Therefore:
\(\frac{AE}{DE} = \frac{CE}{BE} \implies (AE)(BE) = (CE)(DE) \qquad \square\)
Proof Strategy Summary. The key tools are:
- Ratios of sides → look for similar triangles.
- Equal angles → use arc–angle relations (inscribed angles, tangent-chord angles).
- Circles → relate arcs and inscribed/tangent-chord angles.
Problems to Solve for Chapter 17
Problem 294. A point $P$ is outside a circle and is 13 inches from the center. A secant from $P$ cuts the circle at $Q$ and $R$ so that $PQ = 9$ and $QR = 7$. Find the radius of the circle. (AHSME 1954)
Problem 295. Points $A$, $B$, $C$ are on circle $O$. The tangent at $A$ and the secant $BC$ meet at $P$, with $B$ between $C$ and $P$. If $BC = 20$ and $PA = 10\sqrt{3}$, find $PB$. (AHSME 1956)
Problem 296. In the diagram, $EB$ bisects $CD$ and $C$ is the midpoint of $AD$. Find $GB$ if $AB = 16$, $EF = 4$, and $FB = 6$. (MAΘ 1990)
Problem 297. Two tangents are drawn to a circle from exterior point $A$; they touch at $B$ and $C$. A third tangent intersects $AB$ at $P$ and $AC$ at $R$, touching the circle at $Q$. If $AB = 20$, find the perimeter of $\triangle APR$. (AHSME 1961)
Problem 298. A circle is inscribed in a triangle with sides 8, 13, and 17. Let the segments of the side of length 8 made by a point of tangency be $r$ and $s$ with $r < s$. Find $r : s$. (AHSME 1964)
Problem 299. In the figure, center of circle is $O$. $AB \perp BC$, $ADOE$ is a straight line, $AP = AD$, and $AB$ has length twice the radius. Show that $AP^2 = (PB)(AB)$. (AHSME 1960)
Problem 300. Find the area of the inscribed circle of a triangle with sides 20, 21, and 29. (MAΘ 1990)
Problem 301. $AB$ is tangent at $A$ to circle with center $O$; $D$ is interior to the circle; $DB$ intersects the circle at $C$. If $BC = DC = 3$, $OD = 2$, and $AB = 6$, find the radius. (AHSME 1976)
Problem 302. Prove that if quadrilateral $ABCD$ is orthodiagonal (diagonals perpendicular) and circumscribed around a circle, then $(AB)(CD) = (BC)(AD)$. (M&IQ 1991)