Chord distances and equal tangent lengths
Use perpendicular distances, chord bisectors and tangents from one external point with reasons.
Extended only.
Before you begin
- Identify chords, radii and tangents.
- Solve linear equations and use Pythagoras in a right triangle.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A 10 cm chord is bisected. What length is each half?
What you will learn
- Relate equal chords to equal perpendicular distances from the centre.
- Use a chord's perpendicular bisector passing through the centre.
- Solve length problems using equal tangents from one external point.
Measure chord distance perpendicularly
In one circle, equal chords are equidistant from the centre. Distance from a point to a line means the shortest, perpendicular distance, not a sloping line to a chord endpoint. Conversely, chords at equal perpendicular distances from the same centre have equal lengths. Do not compare chords in different-sized circles using this rule without further information.
The perpendicular bisector of a chord passes through the centre. Equivalently, the perpendicular from the centre to a chord bisects it. For radius 13 cm and chord length 10 cm, half the chord is 5 cm. The centre-to-midpoint distance d lies in a right triangle with hypotenuse 13, so d² = 13² − 5² = 144 and d = 12 cm. This is a theorem application, not a required compass-bisector construction.
Tangents from a common point match
If P is outside a circle and PT and PU touch it at T and U, then PT = PU. The tangents must come from the same external point; unrelated tangent segments do not automatically match. Each contact radius is perpendicular to its tangent, which helps identify the right triangles in a diagram.
If PT = 3x + 2 and PU = 5x − 6, set 3x + 2 = 5x − 6. Solving gives x = 4 and both lengths are 14 units. Verify both expressions are positive and equal. If a question asks for a perimeter, include all boundary pieces rather than using the common tangent length as the whole perimeter.
Pause and explain
Equal chords in the same circle have equal what?
Worked example
Two tangents from P have lengths PT = 3x + 2 and PU = 5x − 6. Find x and their lengths.
Show the worked solution
- Tangents from the same external point are equal, so 3x + 2 = 5x − 6.
- Rearrange: 8 = 2x, giving x = 4.
- PT = 3(4) + 2 = 14 and PU = 5(4) − 6 = 14; both are positive.
Answer x = 4; PT = PU = 14 units.
From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A circle has radius 10 cm and a chord 12 cm long. Find the perpendicular distance from its centre to the chord.
Give me a hint
The perpendicular bisects the chord, giving a half-length of 6 cm.
Compare my reasoning
- Use a right triangle with hypotenuse 10 cm and one leg 6 cm.
- The other leg satisfies d² = 10² − 6² = 64.
- Take the positive root: d = 8 cm.
8 cm.
Look for these in your work
- Used half the chord, not the whole chord.
- Recognised the radius as hypotenuse.
Two tangents from P have lengths 2x + 7 and 5x − 5. Find x and the common length.
Give me a hint
Their common external point justifies equality.
Compare my reasoning
- Set 2x + 7 = 5x − 5.
- 12 = 3x, so x = 4.
- Both lengths are 15 units: 2(4) + 7 = 5(4) − 5.
x = 4; common length 15 units.
Look for these in your work
- Gave the equal-tangents reason.
- Checked both expressions and positivity.
Two chords in a circle of radius 13 cm are each 12 cm from the centre, measured perpendicularly. Find each chord length and explain why they match.
Give me a hint
Find half a chord with Pythagoras, then double it.
Compare my reasoning
- Half-chord squared is 13² − 12² = 25.
- The half-length is 5 cm, so each full chord is 10 cm.
- They are equal because equal perpendicular distances from the same centre give equal chords.
Each chord is 10 cm.
Look for these in your work
- Used perpendicular distances in one circle.
- Doubled the half-length and supplied the theorem reason.
Common mistakes
- Using a sloping distance instead of the perpendicular distance.
- Using the full chord as a right-triangle leg.
- Equating tangents that do not share an external point.
What can you explain now?
Where does the perpendicular bisector of a chord pass?
Compare with the explanation
Through the centre of the circle.
The centre is equally distant from the chord's endpoints, so it lies on their perpendicular bisector.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Tomorrow, solve one equal-tangent equation and one chord-distance problem. State the geometric condition before the arithmetic.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.
Original StudyForA teaching content · AI-assisted checks · Human teacher review pending.
Number contains 19 sequenced lessons; Algebra and graphs 21; Coordinate geometry 10; Geometry 15; Mensuration 13; Trigonometry 13; Transformations and vectors 11; Probability 7; Statistics 13. All 72 syllabus section entries now link to teaching and staged practice. Seven mixed checkpoints sample skills and suggest review lessons. Nine earlier overviews remain available. Full cumulative assessment and human teacher review remain pending. Written lesson practice, charts and constructions are self-checked, not automatically graded. Checkpoints do not save an exam grade or certify mastery.
Check the official syllabus 2025–2027 Open related practice and resources