Curriculum Hub → Maths Lessons → Algebra → Graphical functions→ Circles and Chords
Ready-to-teach A-Level lesson. Three video tutorials, worked examples and a free printable worksheet on the equation of a circle and chords.
Chords are the bridge between GCSE circle theorems and A-Level coordinate geometry. Once students can find the midpoint and perpendicular bisector of a chord, they can locate a circle’s centre, derive its equation, and handle tangent problems in Edexcel, AQA and OCR Pure 1.
This page covers the equation of a circle, chord length, and the perpendicular bisector of a chord, with three video tutorials, worked examples and a free printable worksheet.
For teachers: a ready-to-deliver lesson on circles and chords. Editable PowerPoint, differentiated student worksheet (support to challenge) and fully worked solutions. Suitable for classroom delivery, cover lessons or flipped learning. Download the School Flyer
Lesson structure: starter recap → main teaching slides → interactive worked examples → plenary consolidation.
Chord (definition): a chord is a straight line segment whose two endpoints both lie on the circumference of a circle.
The diameter is the longest possible chord, because it passes through the centre. A chord that does not pass through the centre divides the circle into a major segment and a minor segment.
Three chord properties do most of the work at A-Level:
The second property is the one students use most. If you know two points on a circle, the centre must lie somewhere on the perpendicular bisector of the chord joining them. Combine that with one more condition and the centre is fixed.
Equation of a circle — centre (a, b), radius r:(x − a)² + (y − b)² = r²
Length of a chord — where d is the perpendicular distance from the centre to the chord:L = 2√(r² − d²)
Midpoint of a chord joining (x₁, y₁) and (x₂, y₂):M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Perpendicular gradient — if the chord has gradient m, its perpendicular bisector has gradient:m⊥ = −1/m
Distance from a point to a line Ax + By + C = 0:d = |Ax₀ + By₀ + C| / √(A² + B²)
Problem: A and B are points on a circle, where A(2, 3) and B(6, 7). Find the equation of the perpendicular bisector of chord AB.
Teaching point: the centre of the circle must lie somewhere on this line. That single fact unlocks Example 2.
Problem: A circle passes through A(2, 3) and B(6, 7). Its centre lies on the x-axis. Find the equation of the circle.
Always verify with the second point. It catches sign errors in the perpendicular gradient immediately.
Problem: A circle has radius 13. A chord is 5 units from the centre. Find the chord length.
Solution: L = 2√(13² − 5²) = 2√(169 − 25) = 2√144 = 24
Problem: Find the length of the chord where the line 3x + 4y = 20 cuts the circle x² + y² = 25.
Extension: show that 3x + 4y = 25 gives d = 5 = r, so that line is a tangent, not a chord.
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Exam-style questions on chord length, perpendicular bisectors and finding the equation of a circle, with space for working and fully worked solutions.
Three tutorials in sequence, each building on the last. Use them for flipped learning, revision, or as modelled examples on the board.
Skills covered: calculating the midpoint of a line segment; finding the equation of a chord; using coordinate geometry to verify geometric properties.
Skills covered: gradient of a line segment; equation of the perpendicular bisector; proving the bisector passes through the circle’s centre.
Skills covered: the tangent–radius property; deriving the centre and radius from a tangent and chord; forming the equation of a circle under multiple conditions.
| Error | Why it happens | Fix |
|---|---|---|
| Using the chord’s gradient instead of the perpendicular gradient | Students find m and stop | Write “m⊥ = −1/m” as a separate line every time |
| Sign error in (x − a)² + (y − b)² = r² | Negative coordinates in the centre | Centre (−3, 4) gives (x + 3)² + (y − 4)²; substitute a point to check |
| Leaving the answer as r² when asked for the radius | Rushing the final line | Underline what the question asks for before starting |
| Forgetting to double in L = 2√(r² − d²) | Only half the chord is in the triangle | Draw the right-angled triangle every time |
| Assuming the perpendicular bisector passes through the origin | Confusing centre with origin | Emphasise the centre is unknown until it is calculated |
A chord is a straight line segment joining two points on the circumference of a circle. The diameter is the longest chord because it passes through the centre.
If a circle has radius r and the perpendicular distance from the centre to the chord is d, the chord length is L = 2√(r² − d²).
Yes. The perpendicular bisector of any chord always passes through the centre of the circle. This is why two chords are enough to locate the centre.
Find the midpoint and gradient of the chord, then form the perpendicular bisector. The centre lies on that line. Use one further condition such as a second chord, a tangent or a coordinate constraint to fix the centre, then find r² using the distance to a known point.
Yes. A diameter is a chord that passes through the centre, and it is the longest chord in any circle.
Editable PowerPoint, differentiated worksheets and worked solutions, plus 800+ further KS3, GCSE and A-Level resources.
Ideal for: whole-class teaching, department schemes of work and cover lessons.
Instant download · Editable PowerPoint · Differentiated worksheets · Answer keys
Students explore the relationship between a chord and its perpendicular bisector to find the equation of a circle.
As learning progresses, they find the equation of a circle using a chord and tangent.
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