Segment Products and Power of a Point

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Learning goals

Identify the segments in an inside or outside product using endpoint order
Derive the three segment product rules from correctly matched similar triangles
Distinguish signed power from positive products inside, outside, and on a circle
Calculate missing lengths using chord, secant, and tangent products
Check algebraic roots against every actual length and the required point order

Products inside the circle

Multiplying the two parts of a chord through a fixed interior point gives the same result for every such chord. Throughout this lesson, symbols such as $PA$ mean ordinary positive lengths when the endpoints differ.

Let chords $AB$ and $CD$ meet at $P$ , strictly inside one circle. The orders are $A,P,B$ and $C,P,D$ . For lengths $3$ and $8$ on the first chord, and $4$ and $6$ on the second, each product is $24$ .

P lies between the two endpoints of each chord.

Join $AC$ and $DB$ . The angles at $P$ are vertical angles. The angles $PAC$ and $PDB$ intercept the same arc $BC$ , so they are equal by Inscribed Angles. This gives AA similarity, using the criteria from Similarity and Proportion.

Intersecting-chord products from similarity#

First use a different numerical instance: the lengths from $P$ to $A,B,C,D$ are $2,12,3,8$ . The angle equalities just described match triangle $APC$ to triangle $DPB$ . Thus the ratios of corresponding sides give

\begin{aligned} \frac{2}{8}&=\frac{3}{12},\\ 2(12)&=3(8). \end{aligned}

Auxiliary chords reveal the ordered triangle correspondence.

Now keep the same point orders but allow any lengths. Because $P$ lies between the endpoints on both chords, $\angle PAC$ is $\angle BAC$ , and $\angle PDB$ is $\angle CDB$ . Both intercept arc $BC$ excluding their vertices. Also $\angle APC$ and $\angle DPB$ are vertical angles.

AA gives the ordered correspondence

\triangle APC\sim\triangle DPB.

Here $A$ matches $D$ , $P$ matches $P$ , and $C$ matches $B$ . In particular, $PA$ matches $PD$ , and $PC$ matches $PB$ .

\begin{aligned} \frac{PA}{PD}&=\frac{PC}{PB},\\ PA\cdot PB&=PC\cdot PD. \end{aligned}

Nothing in the angle reasoning depended on the numerical lengths. The two chords must be distinct and meet strictly inside; then the triangles are nondegenerate and all denominators are positive.

This is the intersecting-chord product rule. Pair the two lengths on the same straight chord, not two adjacent pieces from different chords.

Worked example 1 Find a part, then the whole chord

Problem. Chords $AB$ and $CD$ intersect at interior point $P$ , with orders $A,P,B$ and $C,P,D$ . The lengths from $P$ to $A,B,C$ are $3$ cm, $10$ cm, and $5$ cm, respectively. Find $PD$ and $CD$ .

P lies between the two endpoints of each chord.

Solution. The two parts of each chord form a product.

\begin{aligned} 3(10)&=5(PD),\\ PD&=6\text{ cm}. \end{aligned}

Point $P$ lies between $C$ and $D$ , so add the parts.

\begin{aligned} CD&=5+6,\\ CD&=11\text{ cm}. \end{aligned}

Both parts are positive, and their product is $30$ , matching the first chord.

Check your understanding

Chords $AB,CD$ meet at interior $P$ . The lengths $PA,PB,PC$ are $4,15,5$ , respectively. Find $PD$ .

3

12

20

60

Products outside the circle

From an external point, the useful pair is the external part and the whole secant segment. Suppose two secant rays meet the same circle in orders $P,A,B$ and $P,C,D$ .

The external segment is $PA$ . The whole secant segment is $PB$ , which includes both $PA$ and the chord $AB$ . For example, an external part of $3$ and an inside chord of $5$ give a whole segment of $8$ . Its product with the external part is $24$ , not $15$ .

Both lengths in each outside product start at P.

Two-secant products from similarity#

For a numerical instance, take $PA$ of $2$ , $PB$ of $15$ , $PC$ of $3$ , and $PD$ of $10$ . Join $AD$ and $CB$ . The two triangles share their angle at $P$ . Their angles at $D$ and $B$ intercept the same arc $AC$ . Therefore triangle $PAD$ matches triangle $PCB$ , and corresponding sides give

\begin{aligned} \frac{2}{3}&=\frac{10}{15},\\ 2(15)&=3(10). \end{aligned}

The common angle is at P, and the inscribed angles at D and B intercept arc AC.

For arbitrary lengths in those same orders, $\angle APD$ and $\angle CPB$ use the same two rays. At $D$ , ray $DP$ passes through $C$ , so $\angle PDA$ is $\angle CDA$ . At $B$ , ray $BP$ passes through $A$ , so $\angle PBC$ is $\angle ABC$ . These inscribed angles intercept the same arc $AC$ excluding $B,D$ .

AA gives

\triangle PAD\sim\triangle PCB.

The matches are $P$ to $P$ , $A$ to $C$ , and $D$ to $B$ .

\begin{aligned} \frac{PA}{PC}&=\frac{PD}{PB},\\ PA\cdot PB&=PC\cdot PD. \end{aligned}

The numerical argument used this same correspondence. The hypotheses are two distinct secant rays from one strict external point to one circle. Each whole segment must be longer than its external part.

This is the two-secant product rule. If a problem gives the external part and the inside chord, first add them.

PB=PA+AB.

Worked example 2 Build both whole secant segments

Problem. Two secants meet a circle in orders $P,A,B$ and $P,C,D$ . The external lengths $PA,PC$ are $3$ cm and $4$ cm. Chord $AB$ is $17$ cm. Find $PD$ and chord $CD$ .

Both lengths in each outside product start at P.

Solution. On the first secant, add the external and inside lengths.

\begin{aligned} PB&=3+17,\\ PB&=20\text{ cm}. \end{aligned}

Use external times whole on each ray.

\begin{aligned} 3(20)&=4(PD),\\ PD&=15\text{ cm}. \end{aligned}

Remove the second external part from its whole.

\begin{aligned} CD&=15-4,\\ CD&=11\text{ cm}. \end{aligned}

The whole length $15$ exceeds the external length $4$ , as the order requires.

Check your understanding

Secants have orders $P,A,B$ and $P,C,D$ . The lengths $PA,AB,PC$ are $2,19,3$ . Find the whole segment $PD$ .

\frac{38}{3}

11

42

14

A tangent and a secant

The square of a tangent segment’s length equals external times whole on a secant from the same strict external point. For example, external length $2$ and whole length $8$ give product $16$ . A tangent segment from that point has length $4$ , whose square is $16$ .

PT touches at T; A is the near secant endpoint and B the far endpoint.

The tangent-secant product from similarity#

In a numerical instance, let tangent segment $PT$ have length $6$ , while secant lengths $PA,PB$ are $3,12$ . Join $TA$ and $TB$ . The tangent-chord theorem matches the angle $PTA$ to angle $PBT$ , using arc $TA$ that excludes $B$ . The angle at $P$ is shared, giving triangle $PTA$ similar to triangle $PBT$ .

\begin{aligned} \frac{6}{12}&=\frac{3}{6},\\ 6^2&=3(12). \end{aligned}

Join TA and TB. The tangent ray TP selects the arc TA excluding B.

For any tangent at $T$ and genuine secant in order $P,A,B$ , the same angle reasoning applies. Ray $TP$ , together with chord $TA$ , selects the arc $TA$ excluding $B$ . By the tangent-chord theorem, its angle equals the inscribed angle $TBA$ , which is also $\angle PBT$ . The angles $TPA$ and $BPT$ use the same two rays.

AA gives

\triangle PTA\sim\triangle PBT.

The matches are $P$ to $P$ , $T$ to $B$ , and $A$ to $T$ .

\begin{aligned} \frac{PT}{PB}&=\frac{PA}{PT},\\ PT^2&=PA\cdot PB. \end{aligned}

All three lengths are positive because $P$ is strictly outside and $A,B$ are distinct. This proof uses the tangent-chord theorem, rather than treating a nearly tangent secant drawing as proof.

Worked example 3 Recover the tangent length

Problem. Segment $PT$ is tangent to a circle at $T$ . A secant from $P$ meets it in order $P,A,B$ , with external length $PA$ of $7$ cm and inside chord $AB$ of $21$ cm. Find $PT$ .

PT touches at T; A is the near secant endpoint and B the far endpoint.

Solution. The whole secant is $28$ cm. The tangent-secant rule gives

\begin{aligned} PT^2&=7(28),\\ PT^2&=196,\\ PT&=14\text{ cm}. \end{aligned}

Although the equation for a real number squared has roots $14$ and $-14$ , $PT$ itself is a length and must be positive.

Check your understanding

The ordered similarity is $\triangle PTA\sim\triangle PBT$ . Its matches are P to P, T to B, and A to T. Which product equality follows from corresponding sides?

PT^2=PA\cdot PB

PT^2=PA\cdot AB

PT=PA\cdot PB

PT^2=AB\cdot PB

One signed value for a point

The three products belong to one idea, but the inside case needs a sign. For a circle with center $O$ and positive radius $r$ , define the signed power of $P$ by

\operatorname{Pow}(P)=OP^2-r^2.

For radius $6$ and center distance $4$ , this signed value is $-20$ . The positive product of two chord parts is $20$ . For the same radius and center distance $10$ , signed power is $64$ , equal to an external-times-whole product.

The link comes from a diameter, without a coordinate equation.

Why the products equal signed power or its negative#

First take radius $5$ and center distance $3$ . The diameter through $O,P$ has endpoints $U,V$ , with $P$ between them. The distances from $P$ to the ends are $8$ and $2$ , so their product is $16$ . The signed value is $3^2-5^2$ , or $-16$ .

The diameter parts have lengths radius plus center distance and radius minus center distance.

For arbitrary radius $r$ and interior center distance $d$ , the diameter parts are $r+d$ and $r-d$ . The chord rule makes every chord product equal to their product.

\begin{aligned} PA\cdot PB&=(r+d)(r-d),\\ PA\cdot PB&=r^2-d^2,\\ PA\cdot PB&=-\operatorname{Pow}(P). \end{aligned}

If $P$ is the center, choose any diameter: both parts are $r$ . This also gives the displayed result with $d$ zero.

Next take radius $5$ and exterior center distance $13$ . Along the diameter ray toward $O$ , the external and whole lengths are $8$ and $18$ . Their product is $144$ , which also equals $13^2-5^2$ .

The near and far endpoints on a diameter give external and whole lengths.

For arbitrary exterior center distance $d$ , the diameter ray has external length $d-r$ and whole length $d+r$ . The two-secant rule compares any other secant with this one.

\begin{aligned} PA\cdot PB&=(d-r)(d+r),\\ PA\cdot PB&=d^2-r^2,\\ PA\cdot PB&=\operatorname{Pow}(P). \end{aligned}

For a tangent segment, the contact radius is perpendicular to it, as proved in Tangents and Secants. In the numerical exterior case, the right triangle gives a tangent square of $13^2-5^2$ , or $144$ . In general, the Pythagorean theorem gives

\begin{aligned} PT^2+r^2&=d^2,\\ PT^2&=d^2-r^2,\\ PT^2&=\operatorname{Pow}(P). \end{aligned}

Both diameter arguments use differences of squares; the choice of numerical radius did not matter.

Position of $P$	Signed power	Positive-length interpretation
Strictly inside	Negative	Chord-part product is its negative
On the circle	Zero	A secant product has a zero factor at $P$
Strictly outside	Positive	External times whole, or the tangent length squared

At a point on the circle, the center distance equals the radius. A secant through that point has one intersection at $P$ itself, so that factor is zero. This is a boundary case, not the strict external configuration: there is one tangent line at $P$ , and no positive segment from $P$ to a different tangent contact point.

Compare products through an interior point

For a fixed circle and a fixed interior point, rotating a chord changes its two parts but preserves their product. Measurements or a moving drawing illustrate the rule; the similarity argument proves it.

Try all five chord directions. Predict which lengths change and whether their positive product changes, then compare signed power separately.

Positive chord products and signed power

Chord

The radius is 4 units. OP = 2 units. PA ≈ 4.61 units. PB ≈ 2.61 units. PA × PB = 12 square units. The signed power of P is -12 square units.

For this fixed interior point, PA and PB are positive segment lengths. Their product is 12; the signed power is negative 12.

Compare products through an exterior point

Keep the exterior point and circle fixed. Different genuine secants have different external and whole lengths but the same product. A drawing supplies evidence for this invariance; the ordered similarity proof establishes it.

Try all five secant directions through fixed $P$ . Track the external length $PA$ and whole length $PB$ , then test whether their product stays fixed.

Use the external segment and the whole secant

Secant

The radius is 4 units. OP = 7 units. PA = 3 units. PB = 11 units. PA × PB = 33 square units. The signed power of P is 33 square units.

Points are ordered P, A, B. The whole length PB includes the external segment PA and the inside segment AB. The external times whole product stays 33.

Worked example 4 Separate the sign from the lengths

Problem. A circle has radius $11$ units. Use the following separate points.

Case A: $OP$ is $7$ units. Find signed power and the positive product for a chord through $P$ .

Case B: $OQ$ is $13$ units. Find signed power and the tangent length from $Q$ .

Case C: $OR$ is $11$ units. Find signed power and describe the tangent at $R$ .

Solution. Case A is inside.

\begin{aligned} \operatorname{Pow}(P)&=7^2-11^2,\\ \operatorname{Pow}(P)&=-72. \end{aligned}

The positive chord product is $72$ square units.

Case B is outside.

\begin{aligned} \operatorname{Pow}(Q)&=13^2-11^2,\\ \operatorname{Pow}(Q)&=48. \end{aligned}

The tangent length is the positive square root, $4\sqrt3$ units.

Case C is on the circle. Its power is zero. There is one tangent line at $R$ ; there are not two positive tangent contact segments starting at $R$ .

Case A: OP is shorter than the radius. The chord-part product is positive 72, while signed power is negative 72.

Case B: Radius OT is perpendicular to tangent QT. The tangent square equals the positive signed power, 48.

Case C: One tangent line is perpendicular to OR at R. Signed power is zero.

Check your understanding

A circle has radius $8$ and $OP$ is $4$ . Which pair gives signed power and the positive chord-part product at $P$ ?

Power $48$

Product $48$

Power $48$

Product $-48$

Power $-48$

Product $48$

Power $-48$

Product $-48$

Algebra must fit the lengths

A root solves an equation; a geometric solution also makes every actual length positive and preserves the stated order. For an external secant, the whole must exceed the external part. If a variable itself names a length, it must be positive. If it is only a parameter inside length expressions, a negative value can be valid.

Worked example 5 Keep a valid negative parameter

Problem. From a strict external point $P$ , a secant meets a circle in order $P,A,B$ . In centimeters, the external length is $x+5$ and the whole length is $x+7$ . A tangent segment from $P$ has length $2\sqrt2$ cm. Find every geometrically valid $x$ , then find $AB$ . Here $x$ is a parameter, not itself a length.

PT touches at T; A is the near secant endpoint and B the far endpoint.

Solution. The tangent square is $8$ . Use external times whole.

\begin{aligned} (x+5)(x+7)&=8,\\ x^2+12x+27&=0,\\ (x+3)(x+9)&=0. \end{aligned}

The algebraic candidates are $-3$ and $-9$ . Check them separately.

For $x=-3$ , the external length is $2$ cm and the whole is $4$ cm. Both are positive, the whole is longer, and their product is $8$ . This negative parameter is valid.

For $x=-9$ , the expressions give $-4$ cm and $-2$ cm. These are not ordinary lengths, so this candidate is invalid.

Only $x=-3$ fits. The inside chord is

\begin{aligned} AB&=4-2,\\ AB&=2\text{ cm}. \end{aligned}

Sometimes more than one algebraic candidate passes every length and order check. Keep every admissible candidate; the sign of a parameter alone cannot decide.

Check your understanding

A secant has external length $x+4$ and whole length $x+10$ . A tangent from the same point has length $3\sqrt3$ . Here $x$ is a parameter. Which value is geometrically valid?

x=-13

x=-1

C Both

-1

and

-13

D Neither value

Segment Products and Power of a Point

Learning goals

Products inside the circle

Products outside the circle

A tangent and a secant

One signed value for a point

Compare products through an interior point

Compare products through an exterior point

Algebra must fit the lengths

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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