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Segment Products and Power of a Point: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Recover a chord from its parts

Chords $AB$ and $CD$ of one circle meet at $P$ , with orders $A,P,B$ and $C,P,D$ . The lengths from $P$ to $A,B,C$ are $7$ cm, $8$ cm, and $4$ cm, respectively.

Find $PD$ and the full chord $CD$ .

The orders are A, P, B and C, P, D.

Text description of this figure

Chords have orders A, P, B and C, P, D inside one circle. This schematic preserves incidence and endpoint order. Use the lengths in the statement, not the proportions drawn.
1. Hint 1
  Use the two opposite parts on the same chord as one pair.
2. Hint 2
  Find the missing part from the shared product, then use the inside point order to recover the full chord.
Answer

$\begin{gathered}PD=14\text{ cm},\\CD=18\text{ cm}.\end{gathered}$

Full solution

The product of the parts on AB equals the product of the parts on CD.
$\begin{aligned}7(8)&=4(PD),\\PD&=14\text{ cm}.\end{aligned}$

Because P lies between C and D, the full chord is the sum of its parts.
$\begin{aligned}CD&=4+14,\\CD&=18\text{ cm}.\end{aligned}$

Both parts are positive, and each chord gives product 56 square centimeters.

Answer

$\begin{gathered}PD=14\text{ cm},\\CD=18\text{ cm}.\end{gathered}$

Key idea

An inside product finds a part before addition gives the whole chord.
Problem 2 Outside parts and whole segments

Two secants from external $P$ meet one circle in orders $P,A,B$ and $P,C,D$ . The lengths $PA,AB,PC$ are $2$ cm, $16$ cm, and $3$ cm, respectively.

Find the whole length $PD$ and the inside chord $CD$ .

A and C are the near endpoints; B and D are the far endpoints.

Text description of this figure

Two secants have orders P, A, B and P, C, D. This schematic preserves incidence and endpoint order. Use the lengths in the statement, not the proportions drawn.
1. Hint 1
  Each product outside uses two distances starting at P.
2. Hint 2
  Build the first whole secant by addition, find the second whole from its product, and then remove its external part.
Answer

$\begin{gathered}PD=12\text{ cm},\\CD=9\text{ cm}.\end{gathered}$

Full solution

The first whole secant has length 18 cm.
Use external times whole.
$\begin{aligned}2(18)&=3(PD),\\PD&=12\text{ cm}.\end{aligned}$

The inside chord is the second whole minus its external part.
$\begin{aligned}CD&=12-3,\\CD&=9\text{ cm}.\end{aligned}$

Each whole exceeds its external part, and each product is 36 square centimeters.

Answer

$\begin{gathered}PD=12\text{ cm},\\CD=9\text{ cm}.\end{gathered}$

Key idea

Identify the whole secant before multiplying.
Problem 3 From a tangent to an inside chord

Segment $PT$ is tangent at $T$ . A secant from the same strict external point $P$ meets the circle in order $P,A,B$ . The tangent length is $15$ cm, and the external length $PA$ is $9$ cm.

Find $PB$ and $AB$ .

PT touches at T and the secant passes through A before B.

Text description of this figure

PT is tangent at its actual contact T. The secant has order P, A, B. This schematic preserves incidence and endpoint order. Use the lengths in the statement, not the proportions drawn.
1. Hint 1
  The tangent square gives the product for the secant.
2. Hint 2
  Divide that product by the external length to find the whole, then subtract the external part.
Answer

$\begin{gathered}PB=25\text{ cm},\\AB=16\text{ cm}.\end{gathered}$

Full solution

The tangent-secant rule uses the whole segment PB.
$\begin{aligned}15^2&=9(PB),\\PB&=25\text{ cm}.\end{aligned}$

A lies between P and B.
$\begin{aligned}AB&=25-9,\\AB&=16\text{ cm}.\end{aligned}$

The positive lengths satisfy the required order, and the secant product is 225 square centimeters.

Answer

$\begin{gathered}PB=25\text{ cm},\\AB=16\text{ cm}.\end{gathered}$

Key idea

A tangent square determines the external-times-whole product.
Problem 4 Check the first invalid conclusion

A tangent segment from external $P$ has length $2\sqrt6$ cm. A secant has external length $x$ cm and inside chord length $5$ cm. Here x itself is a length.

The following work is proposed.

Line 1: $x(x+5)=24$ .

Line 2: $(x-3)(x+8)=0$ .

Line 3 gives both candidates:
$\begin{gathered}x=3,\\x=-8.\end{gathered}$
Line 4: The geometry has two possible external lengths, 3 cm and -8 cm.

Identify the first invalid line, correct the conclusion, and give the whole secant length.

The external segment and whole segment are both ordinary lengths.

Text description of this figure

PT is tangent at its actual contact T. The secant has order P, A, B. This schematic preserves incidence and endpoint order. Use the lengths in the statement, not the proportions drawn.
1. Hint 1
  Distinguish solving an equation from obtaining a possible length.
2. Hint 2
  Substitute each candidate into both the external and whole length expressions.
Answer

First invalid line: Line 4. The only external length is $3$ cm; the whole secant length is $8$ cm.

Full solution

Line 1 is correct: the tangent square is 24, and the whole secant is x+5.

Line 2 is correct because expanding the factors gives the equation from Line 1 with 24 subtracted.
Line 3 correctly lists both algebraic roots.

Line 4 is the first invalid conclusion.
At x=-8 the external length is -8 cm and the whole is -3 cm.
Ordinary lengths cannot be negative.

At x=3, the external length is 3 cm and the whole length is 8 cm.
Both are positive and correctly ordered.
Their product is 24, matching the tangent square.

Answer

First invalid line: Line 4. The only external length is $3$ cm; the whole secant length is $8$ cm.

Key idea

An algebraically correct root list can still contain geometrically invalid candidates.
Problem 5 Read a signed value inside

A circle has center $O$ and radius $10$ cm. Point $P$ is $\sqrt{37}$ cm from $O$ . A chord through $P$ has endpoints in order $A,P,B$ , with $PA$ of $7$ cm.

Find the signed power of $P$ and the length $PB$ . Explain the sign used in the chord product.

The radius, center distance OP, and chord part PA are givens. Points A, P, B lie on one chord in that order.

Text description of this figure

A circle has center O and radius 10 cm. Segment OP has length √37 cm. Chord AB passes through P in the order A, P, B, with PA equal to 7 cm. The radius, OP, and PA labels are the given lengths.
1. Hint 1
  Compare the squared center distance with radius squared.
2. Hint 2
  Inside the circle the positive chord-part product is the negative of signed power.
Answer

$\begin{gathered}\operatorname{Pow}(P)=-63\text{ cm}^2,\\PB=9\text{ cm}.\end{gathered}$
The positive chord product is $63$ square centimeters.

Full solution

The squared center distance is 37, smaller than 100, so P is inside.
$\begin{aligned}\operatorname{Pow}(P)&=37-100,\\\operatorname{Pow}(P)&=-63\text{ cm}^2.\end{aligned}$

Ordinary positive chord lengths have a positive product.
Use the negative of signed power.
$\begin{aligned}7(PB)&=63,\\PB&=9\text{ cm}.\end{aligned}$

The two chord parts are positive; their product is 63 square centimeters, rather than -63.

Answer

$\begin{gathered}\operatorname{Pow}(P)=-63\text{ cm}^2,\\PB=9\text{ cm}.\end{gathered}$
The positive chord product is $63$ square centimeters.

Key idea

The sign belongs to power, not to an ordinary segment length.
Problem 6 A straight support beside a disk

A circular display disk is outlined by a frame. From an external mounting point P, a straight measuring line reaches the near frame point A after 6 cm and then travels 18 cm across the disk to far frame point B.

A straight support from P touches the frame tangentially at T. Find the support length PT, ignoring the thickness of the support and frame.

The frame is the circle; PT is tangent and P, A, B lie in order.

Text description of this figure

PT is tangent at its actual contact T. The secant has order P, A, B. This schematic preserves incidence and endpoint order. Use the lengths in the statement, not the proportions drawn.
1. Hint 1
  Interpret the measured outside part and inside chord before choosing a relation.
2. Hint 2
  The support length squared equals external times whole on the measuring line.
Answer

The support is $12$ cm long.

Full solution

The measuring line has external length 6 cm and whole length 24 cm.

The tangent support meets the same circle from the same external point.
$\begin{aligned}PT^2&=6(24),\\PT^2&=144,\\PT&=12\text{ cm}.\end{aligned}$

The positive root gives the physical support length.

Answer

The support is $12$ cm long.

Key idea

A tangent support can be found from an external and whole secant measurement.
Problem 7 The boundary at a circle point

A circle has radius 17 cm and diameter AB. Point P coincides with A.

Find the signed power of P and the product of its distances to A and B.

State the number of tangent lines through P and whether there is a positive tangent segment from P to a different contact point.
1. Hint 1
  A point on the circle has center distance equal to the radius.
2. Hint 2
  One secant factor is the distance from P to itself. Distinguish a tangent line at P from two positive contact segments.
Answer

Power $0$ ; distance product $0$ ; one tangent line through P; no positive segment from P to a different tangent contact point.

Full solution

P is on the circle, so its power is the difference of equal squared lengths.
$\begin{aligned}\operatorname{Pow}(P)&=17^2-17^2,\\\operatorname{Pow}(P)&=0.\end{aligned}$

P coincides with A, so PA is zero.
PB is the diameter length, 34 cm.
Therefore the distance product is zero.

There is exactly one tangent line at P, perpendicular to the radius there.
Its contact point is P itself.
A different contact point cannot give another tangent through P, so there is no positive tangent contact segment of the requested kind.

Answer

Power $0$ ; distance product $0$ ; one tangent line through P; no positive segment from P to a different tangent contact point.

Key idea

Zero power is an on-circle boundary, not a strict external two-tangent configuration.
Problem 8 Keep both possible parameter values

Chords AB and CD meet at interior P, with orders A, P, B and C, P, D. In centimeters, the four part lengths are given below.

$PA=x+5$ .

$PB=5-x$ .

$PC=4$ .

$PD=4$ .

Here x is a parameter. Find every valid x. For each case, give PA and PB and say which of A and B is nearer P.

Only the endpoint order is represented; either valid labeled-length case may occur.

Text description of this figure

Chords have orders A, P, B and C, P, D inside one circle. This schematic preserves incidence and endpoint order. Use the lengths in the statement, not the proportions drawn.
1. Hint 1
  The equal products give an equation whose candidates may have different signs.
2. Hint 2
  For each candidate, test both first-chord lengths. The nearer endpoint has the smaller positive distance.
Answer

$\begin{gathered}x=3,\\PA=8\text{ cm},\\PB=2\text{ cm}.\end{gathered}$
B is nearer P.
$\begin{gathered}x=-3,\\PA=2\text{ cm},\\PB=8\text{ cm}.\end{gathered}$
A is nearer P.

Full solution

The positive product on CD is 16 square centimeters.
$\begin{aligned}(x+5)(5-x)&=16,\\25-x^2&=16,\\x^2&=9.\end{aligned}$

Case x=3: PA is 8 cm and PB is 2 cm.
Both are positive, their product is 16, and P lies between the endpoints.
B is nearer P.

Case x=-3: PA is 2 cm and PB is 8 cm.
Both are positive and their product is also 16.
A is nearer P.

Both parameter values fit.
A negative parameter is allowed here because x itself does not name a length.

Answer

$\begin{gathered}x=3,\\PA=8\text{ cm},\\PB=2\text{ cm}.\end{gathered}$
B is nearer P.
$\begin{gathered}x=-3,\\PA=2\text{ cm},\\PB=8\text{ cm}.\end{gathered}$
A is nearer P.

Key idea

Retain every candidate that makes all actual lengths positive in the stated order.
Problem 9 A midpoint consequence

From strict external P, a secant meets one circle in order P, A, B and a tangent touches it at T. Let M be the midpoint of chord AB.

Prove that $PM^2-AM^2=PT^2$ . Use ordinary lengths and justify the point order in the subtraction.

The midpoint M is introduced in the statement; it is added to the solution drawing.

Text description of this figure

PT is tangent at its actual contact T. The secant has order P, A, B. This schematic preserves incidence and endpoint order. Use the lengths in the statement, not the proportions drawn.
1. Hint 1
  The midpoint lies inside chord AB, so the external point comes before A and then M.
2. Hint 2
  Express PA and PB as the difference and sum of PM and AM, then use the tangent-secant relation.
Answer

$PM^2-AM^2=PT^2.$
Since P, A, M, B occur in order, both secant factors are positive.

Full solution

Because M is the midpoint of AB and P is external on its extension, the order is P, A, M, B.
In particular, PM exceeds AM.

The equal chord halves give
$\begin{aligned}PA&=PM-AM,\\PB&=PM+AM.\end{aligned}$

Use the tangent-secant product.
$\begin{aligned}PT^2&=(PM-AM)(PM+AM),\\PT^2&=PM^2-AM^2.\end{aligned}$

The sum and difference are positive actual secant lengths in the stated order, so no directed-length convention is needed.

M splits AB into equal halves, so PA is PM minus AM and PB is PM plus AM.

Answer

$PM^2-AM^2=PT^2.$
Since P, A, M, B occur in order, both secant factors are positive.

Key idea

A midpoint can turn the secant product into a difference of squares.
Problem 10 A root, a diameter, and a further length

A secant from external P runs along diameter AB of a circle with center O, in order P, A, O, B. A second secant has order P, C, D.

In centimeters, PA has length x and AB has length 2x+3. Here x itself is a length. On the second secant, PC is 6 cm and CD is 9 cm.

Find the valid x, then find the radius, the center distance OP, and the length of a tangent segment from P.

The first secant passes through the center; the second meets the circle twice.

Text description of this figure

External P has a diameter secant in order P, A, O, B and another secant in order P, C, D. The construction is schematic and has no computed length labels.
1. Hint 1
  Build each whole secant and compare external times whole.
2. Hint 2
  Test both quadratic roots against the actual lengths, including the fact that x itself is a length.
3. Hint 3
  Once the diameter is known, use its midpoint to find the radius and center distance. The shared product also gives the tangent square.
Answer

$\begin{gathered}x=5,\\r=\frac{13}{2}\text{ cm},\\OP=\frac{23}{2}\text{ cm},\\PT=3\sqrt{10}\text{ cm}.\end{gathered}$

Full solution

The second whole secant is 15 cm.
The first whole secant is 3x+3.
$\begin{aligned}x(3x+3)&=6(15),\\x^2+x-30&=0,\\(x-5)(x+6)&=0.\end{aligned}$

Candidate x=5: the external length is 5 cm, the diameter is 13 cm, and the whole secant is 18 cm.
All are positive and the whole exceeds its external part.

Candidate x=-6: the external length is -6 cm, the diameter expression is -9 cm, and the whole expression is -15 cm.
These fail the length conditions.
Only x=5 is valid.

The radius is half the 13 cm diameter.
O lies one radius beyond A.
$\begin{aligned}r&=\frac{13}{2}\text{ cm},\\OP&=5+\frac{13}{2},\\OP&=\frac{23}{2}\text{ cm}.\end{aligned}$

The tangent square equals the shared secant product, 90 square centimeters.
$\begin{aligned}PT^2&=90,\\PT&=3\sqrt{10}\text{ cm}.\end{aligned}$

As a check, the difference of the squared center distance and squared radius is 90.
The second chord is 9 cm, less than the 13 cm diameter, consistent with a genuine second secant.

Answer

$\begin{gathered}x=5,\\r=\frac{13}{2}\text{ cm},\\OP=\frac{23}{2}\text{ cm},\\PT=3\sqrt{10}\text{ cm}.\end{gathered}$

Key idea

After selecting a valid root, carry the point order through every further length.

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