How to Calculate Segment Lengths in Circles Using the Power Theorem

Segment lengths in a circle follow three power theorems: chord-chord, secant-secant and secant-tangent. Each says that two products of segment lengths are equal, so one unknown length can always be solved for. Below you will find all three theorems, worked examples, and practice questions.

Step-by-step Guide: Segment Lengths in Circle

1. Chord-Chord Power Theorem:
When two chords intersect inside a circle, the product of the lengths of the segments of one chord equals the product of the lengths of the segments of the other chord.
Mathematically, if \(AB\) and \(CD\) are intersecting chords, then:
\( AE \times EB = CE \times ED \)

2. Tangent-Secant Power Theorem:
When a tangent and a secant intersect outside a circle, the square of the length of the tangent segment equals the product of the lengths of the secant’s entire length and its external segment.
Mathematically, if \(PA\) is a tangent and \(PBC\) is a secant, then:
\( PA^2 = PB \times PC \)

3. Secant-Secant Power Theorem:
When two secants intersect outside a circle, the product of the lengths of one secant and its external segment equals the product of the lengths of the other secant and its external segment.
Mathematically, if (PAB) and (PCD) are secants that meet outside the circle at P, then:
\( PA \times PB = PC \times PD \)

Examples

Example 1:
Two intersecting chords in a circle, \(AB\) and \(CD\), are divided into segments of \(3 \text{ cm}\), \(4 \text{ cm}\), \(2 \text{ cm}\), and \(x\) respectively. Using the Chord-Chord Power Theorem, determine the value of \(x\).

Solution:
According to the Chord-Chord Power Theorem:
\( AE \times EB = CE \times ED \)
Given: \( AE = 3 \text{ cm} \) and \( EB = 4 \text{ cm} \)
\( 3 \times 4 = 2 \times x \)
\( x = 6 \text{ cm} \)

Example 2:
A circle has a tangent \(PA\) of length \(5 \text{ cm}\) and a secant \(PBC\) with segment \(PB\) measuring \(3 \text{ cm}\) and \(PC\) measuring \(7 \text{ cm}\). Validate the Tangent-Secant Power Theorem with the given data.

Solution:
According to the Tangent-Secant Power Theorem:
\( PA^2 = PB \times PC \)
Given: \( PA = 5 \text{ cm} \), \( PB = 3 \text{ cm} \), \( PC = 7 \text{ cm} \)
\( 5^2 = 3 \times 7 \)
These lengths cannot come from a real tangent and secant, since \(25 \neq 21\).

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Example 3:
Two secants, \(PAB\) and \(PCD\), intersect outside a circle. The given measurements are: \(PA = 2 \text{ cm}\), \(AB = 4 \text{ cm}\), \(PC = 3 \text{ cm}\), and \(CD = y\). Use the Secant-Secant Power Theorem to determine the value of \(y\).

Solution:
According to the Secant-Secant Power Theorem:
\( PA \times PB = PC \times PD \)
Given: \( PB = PA + AB = 2+4 = 6 \text{ cm} \), \( PD = 3 + y \text{ cm} \)
\( 2 \times 6 = 3 \times (3+y) \)
On solving, \( y = 1 \text{ cm} \).

Practice Questions:

1. Two intersecting chords in a circle have lengths of segments as \(2 \text{ cm}\), \(8 \text{ cm}\), \(3 \text{ cm}\), and \(x\). Find \(x\).
2. A tangent of length \(6 \text{ cm}\) intersects with a secant of full length \(10 \text{ cm}\) and what is the external segment?

Answers:

1. \( x = 5.33 \text{ cm} \)
2. External segment \( = 3.6 \text{ cm} \)

The Three Power Theorems at a Glance

All three come from one idea called the power of a point: for a fixed point and a fixed circle, a certain product of distances is the same no matter which line you draw through the point.

Key words: the outside part (external segment) runs from the outside point to the first place the secant hits the circle. The whole secant runs from the outside point to the far side of the circle.

Worked Example With a Quadratic

Two secants leave point P. On the first, the outside part is 4 and the part inside the circle is 6. On the second, the outside part is x and the inside part is 3. Find x.

1. Whole first secant: 4 + 6 = 10. Whole second secant: x + 3.
2. Set up the theorem: 4 × 10 = x(x + 3).
3. Simplify: x² + 3x − 40 = 0.
4. Factor: (x + 8)(x − 5) = 0, so x = 5 or x = −8.
5. A length can’t be negative, so x = 5. Check: 5 × 8 = 40.

Common mistake: multiplying the outside part by the inside part. Here 4 × 6 = 24 would give the wrong answer. Always use the whole secant.

Frequently Asked Questions

How do you find segment lengths in circles?

First decide where the lines meet. If two chords cross inside the circle, use the chord-chord theorem. If two secants, or a tangent and a secant, meet outside the circle, use the secant-secant or tangent-secant theorem. Each one sets two products of segment lengths equal, so you write one equation and solve for the unknown length.

What is the chord-chord power theorem?

When two chords intersect inside a circle, the two parts of one chord multiply to the same number as the two parts of the other chord: AE × EB = CE × ED. For example, if one chord is split into 8 and 3 and the other into 4 and x, then 4x = 24, so x = 6.

What is the secant-secant power theorem?

When two secants are drawn from the same outside point, the outside segment times the whole secant is equal for both. If one secant has an outside part of 3 and a whole length of 12, the product is 36. For a second secant with an outside part of 4, the whole length must be 36 ÷ 4 = 9.

What is the tangent-secant power theorem?

If a tangent and a secant are drawn from the same outside point, the tangent length squared equals the outside part of the secant times the whole secant. For a secant with an outside part of 5 and a whole length of 20, the tangent t satisfies t² = 5 × 20 = 100, so t = 10. Use only the positive root, since lengths are positive.

What are the formulas for segment lengths in circles?

There are three. Chord-chord: part × part = part × part, or AE × EB = CE × ED. Secant-secant: outside × whole = outside × whole, or PA × PB = PC × PD. Tangent-secant: tangent² = outside × whole, or PA² = PB × PC. When an unknown appears in more than one segment, these can lead to quadratic equations.

Is the segments of chords theorem the same as the chord-chord power theorem?

Yes. Both names describe the same rule: when two chords intersect inside a circle, the products of their segments are equal. It can be proved with similar triangles, since the angles formed by the crossing chords and the inscribed angles match. Some textbooks call it the intersecting chords theorem, but the equation AE × EB = CE × ED is the same.