Circle theorems
Angle at the centre, angle in a semicircle, same segment, cyclic quadrilaterals, and the tangent-radius right angle
About three lessons of 45 to 60 minutes
Five facts hide inside every circle, and they let you find angles nobody measured
Draw any circle. Pick two points A and B on it. From the centre, the angle looking out at A and B is always EXACTLY double the angle seen from any point P on the far arc, no matter where P slides. Slide P along its arc and its angle refuses to change; move the angle's vertex to the centre and it doubles. These are not coincidences of one drawing; they are theorems, true for every circle ever drawn.
GCSE Higher asks you to do two things with them: APPLY them, chasing an unknown angle through a diagram one theorem at a time while naming each theorem used (the reason earns its own marks), and PROVE them, showing with isosceles triangles built from equal radii why they must be true. This unit builds the five standard facts, then walks the central proof.
- A Ferris wheel pod looking at two gates on the rimthe angle between the gates looks the same from every pod on the far arc: same segment
- A triangle drawn on a diameterthe angle at the far side of the circle is always exactly 90 degrees
- A wheel resting on flat groundthe ground is a tangent, and the spoke to the contact point meets it at exactly 90 degrees
- Any four points on a circle joined in orderopposite corners of the quadrilateral always add to exactly 180 degrees
What students will be able to do
Students will state and apply the standard circle theorems (angle at the centre is twice the angle at the circumference, angle in a semicircle is 90 degrees, angles in the same segment are equal, opposite angles of a cyclic quadrilateral sum to 180 degrees, and a tangent meets a radius at 90 degrees) to find unknown angles with reasons, and will follow and reproduce the radius-isosceles proof of the angle-at-the-centre theorem.
- I can find an angle using 'the angle at the centre is twice the angle at the circumference', in either direction.
- I can use 'the angle in a semicircle is 90 degrees' together with the triangle angle sum.
- I can explain why two angles in the same segment are equal, and use it.
- I can find missing angles of a cyclic quadrilateral using opposite angles summing to 180 degrees.
- I can use the tangent-radius right angle to solve a tangent triangle.
- I always write the theorem's NAME next to the angle it produced.
Standards this unit teaches
- GCSE Geometry and measures #10 (Higher tier)UK GCSE Mathematics (DfE, England)Circle theorems
Subject content statement (Department for Education, "GCSE mathematics: subject content and assessment objectives", published 1 November 2013, reference DFE-00233-2013, "Geometry and measures" section, "Properties and constructions", item 10, https://www.gov.uk/government/publications/gcse-mathematics-subject-content-and-assessment-objectives): students should "apply and prove the standard circle theorems concerning angles, radii, tangents and chords, and use them to prove related results". The entire item is BOLD type: circle theorems are assessed only at the higher tier.
Prior knowledge
This unit builds on skills students should already have met. Revisit any that are shaky first.
- Year 7 Angle Facts teaching unitangles on a line, at a point, and the triangle sum, used in every angle chase here
- Year 11 Circle Sectors teaching unitcircle vocabulary (arc, sector, radius) this unit builds on
- Year 8-9 Congruence & Similarity teaching unitthe isosceles-triangle reasoning the proofs are made of
Words to teach and display
- Chord
- a straight line joining two points on a circle (a diameter is a chord through the centre)
- Tangent
- a straight line that touches the circle at exactly one point
- Arc
- a piece of the circle's edge between two points; the two points cut the circle into a major and a minor arc
- Segment
- the region between a chord and one of its arcs; a chord makes a major and a minor segment
- Subtend
- the arc AB is said to subtend the angle made at a third point by the lines from A and B
- Cyclic quadrilateral
- a four-sided shape whose four corners all lie on one circle
Teach it: concrete, pictorial, abstract
The lesson moves from things students can hold, to pictures and diagrams, to the written maths. The diagrams below are drawn from data, so they are accurate and print cleanly. Teach straight from them.
1. The angle at the centre, and the semicircle special case
ConcreteThe parent theorem: the angle subtended by an arc at the CENTRE of a circle is exactly twice the angle subtended by the same arc at the CIRCUMFERENCE. Every other fact in this unit is a child or cousin of this one.
In the figure, points A, B and P lie on the circle and O is the centre. The arc AB (the lower arc, not containing P) subtends angle AOB = 80 degrees at the centre and angle APB at the circumference. The theorem says angle APB = 80 / 2 = 40 degrees, wherever P sits on the major arc.
Now let the chord AB grow into a DIAMETER. The 'angle at the centre' becomes the straight line AOB, 180 degrees, so the angle at the circumference must be 180 / 2 = 90 degrees. That corollary has its own name: the angle in a semicircle is a right angle. Spotting a diameter in a diagram should trigger an automatic hunt for the hidden right angle.
A, B and P lie on a circle with centre O, with P on the major arc. Angle AOB = 124 degrees. Find angle APB, with a reason.
- Angle AOB (at the centre) and angle APB (at the circumference) stand on the same arc AB.
- The angle at the centre is twice the angle at the circumference.
- So angle APB = 124 / 2 = 62 degrees.
Answer: 62 degrees, because the angle at the centre is twice the angle at the circumference on the same arc.
- If angle APB = 51 degrees, what is angle AOB, and which direction did you use the theorem?
- Why is the angle in a semicircle exactly the angle-at-the-centre theorem applied to a 180-degree 'angle'?
2. Same segment, and cyclic quadrilaterals
PictorialTwo more children of the centre theorem. If two angles at the circumference stand on the same arc, each is half the SAME central angle, so they must equal each other: angles in the same segment are equal. And joining four circle points in order makes a cyclic quadrilateral whose opposite angles always sum to 180 degrees.
Same segment: P and Q both sit on the major arc of chord AB. Angle APB and angle AQB are each half of angle AOB, so angle APB = angle AQB. Sliding the vertex along its arc changes the triangle's shape but never that angle.
Cyclic quadrilateral: for ABCD on a circle, the angles at A and at C stand on the two OPPOSITE arcs BD, whose central angles wrap the whole circle: they sum to 360 degrees. Halving both: angle A + angle C = 180 degrees, and likewise angle B + angle D = 180 degrees. In the figure, 95° + 85° = 180° and 80° + 100° = 180°.
ABCD is a cyclic quadrilateral with angle A = 104 degrees and angle B = 77 degrees. Find angles C and D, with a reason.
- Opposite angles of a cyclic quadrilateral sum to 180 degrees.
- Angle C is opposite angle A: C = 180 - 104 = 76 degrees.
- Angle D is opposite angle B: D = 180 - 77 = 103 degrees.
Answer: C = 76 degrees and D = 103 degrees, because opposite angles of a cyclic quadrilateral are supplementary.
- Why does the same-segment theorem fail if P and Q are on OPPOSITE sides of the chord?
- Can a cyclic quadrilateral have two opposite angles of 100 degrees each? Why not?
3. The tangent-radius right angle, and proving the centre theorem
AbstractA tangent touches the circle at one point T, and the radius OT to that point always meets it at exactly 90 degrees. That right angle turns every tangent diagram into right-triangle work. The section closes with the proof GCSE expects: WHY the centre angle doubles.
In the figure, PT is a tangent touching at T, O is the centre, and angle OPT = 32 degrees. Since angle OTP = 90 degrees (tangent perpendicular to radius), triangle OTP gives angle TOP = 180 - 90 - 32 = 58 degrees.
Proof of the centre theorem (the case with O inside angle APB): draw the line PO and extend it. Triangles OAP and OBP are isosceles, because OA = OP = OB are all radii. In an isosceles triangle the base angles are equal, and an exterior angle of a triangle equals the sum of the two interior opposite angles. So the extended line splits angle AOB into two exterior angles, each DOUBLE its matching half of angle APB. Adding the halves: angle AOB = 2 x angle APB. Every step is a Year 7 angle fact; the circle's only contribution is that all radii are equal.
PT is a tangent to a circle with centre O, touching at T. Angle TOP = 49 degrees. Find angle OPT, with reasons.
- The tangent meets the radius at the point of contact at 90 degrees, so angle OTP = 90 degrees.
- The angles of triangle OTP sum to 180 degrees.
- Angle OPT = 180 - 90 - 49 = 41 degrees.
Answer: 41 degrees, using the tangent-radius right angle and the triangle angle sum.
- Where exactly does the circle-ness enter the proof of the centre theorem? (What fact would fail for a non-circle?)
- A tangent diagram shows angle OTP as 85 degrees. How do you know the diagram is wrong before finding any other angle?
Common misconceptions and how to address them
MisconceptionThe angle at the centre is twice ANY angle at the circumference in the diagram.
Why it happens: Students apply the doubling without checking that both angles stand on the SAME arc AB.
How to address it: Before doubling or halving, trace both angles back to the arc they stand on with a finger (or shade the arc). Only angles standing on the same arc are locked in the 2:1 ratio.
MisconceptionADJACENT angles of a cyclic quadrilateral sum to 180 degrees.
Why it happens: The word 'opposite' gets dropped, and adjacent angles sometimes do happen to sum to 180 in special cases, which seems to confirm the error.
How to address it: Mark opposite vertex pairs (A with C, B with D) before writing any equation, and check: A + C = 180 and B + D = 180, never A + B in general. The four angles still sum to 360 as in any quadrilateral, which is a free sanity check on the finished answer.
MisconceptionWriting only the number ('x = 62') is a full answer to a circle-theorem question.
Why it happens: In earlier angle work the reason was often optional, so students skip it under exam pressure.
How to address it: GCSE circle-theorem questions carry a mark for the NAMED theorem. Train the sentence habit: '62 degrees, because the angle at the centre is twice the angle at the circumference'. The worksheet answer keys in this unit model that phrasing on every item.
Guided practice (with answers)
1. Angle AOB = 96 degrees at the centre, on the same arc as P at the circumference. Find angle APB.
Answer: 48 degrees: the angle at the centre is twice the angle at the circumference.
2. AB is a diameter and P is on the circle with angle PAB = 28 degrees. Find angle PBA.
Answer: 62 degrees: angle APB = 90 (angle in a semicircle), so 180 - 90 - 28 = 62.
3. P and Q lie on the same arc of chord AB, and angle APB = 44 degrees. Find angle AQB.
Answer: 44 degrees: angles in the same segment are equal.
4. In cyclic quadrilateral ABCD, angle A = 91 degrees. Find angle C.
Answer: 89 degrees: opposite angles of a cyclic quadrilateral sum to 180.
5. PT is a tangent touching at T, and angle OPT = 37 degrees. Find angle TOP.
Answer: 53 degrees: angle OTP = 90 (tangent meets radius), so 180 - 90 - 37 = 53.
Independent practice worksheets
Practise all five theorems with accurately drawn, computed diagrams and reason-first answer keys.
Differentiation
- Issue a five-row theorem card (name, one-line statement, thumbnail sketch) and require the row number to be quoted next to every found angle until the names are automatic.
- Colour-code each angle chase: shade the arc an angle stands on before applying the centre or same-segment theorem, so 'same arc' is checked visually.
- Start every tangent question by drawing the right-angle marker at the contact point before anything else is attempted.
- Prove the angle-in-a-semicircle corollary in two lines from the centre theorem.
- Prove that opposite angles of a cyclic quadrilateral are supplementary, using the centre theorem on both arcs BD.
- Investigate the alternate segment theorem (the angle between a tangent and a chord equals the angle in the alternate segment), the sixth standard theorem, and test it by drawing.
Assessment: exit ticket
A three-question exit ticket sampling the centre theorem, cyclic quadrilaterals, and the tangent right angle, reasons required.
1. Angle APB = 58 degrees at the circumference, standing on the same arc as angle AOB at the centre. Find angle AOB, with a reason.
Answer: 116 degrees: the angle at the centre is twice the angle at the circumference.
2. In cyclic quadrilateral ABCD, angle B = 113 degrees. Find angle D, with a reason.
Answer: 67 degrees: opposite angles of a cyclic quadrilateral sum to 180.
3. PT is a tangent touching the circle at T, centre O, and angle TOP = 64 degrees. Find angle OPT, with reasons.
Answer: 26 degrees: the tangent meets the radius at 90 degrees, and the triangle's angles sum to 180.
Teacher notes and timings
- Rough timing: Lesson 1 the centre theorem and semicircle (section 1), Lesson 2 same segment and cyclic quadrilaterals (section 2), Lesson 3 tangents plus the proof and exit ticket (section 3).
- Tier note: this is a HIGHER TIER unit. In DfE reference DFE-00233-2013, Geometry and measures item 10 ('apply and prove the standard circle theorems concerning angles, radii, tangents and chords, and use them to prove related results') is entirely bold type, which the document's type key defines as content assessed only at the higher tier.
- Every figure in this unit is drawn by the CircleTheoremFigure engine from the stated angle values themselves (point positions are derived via the inscribed-angle theorem, and the matching worksheet items' drawn configurations are re-measured by tests/ukgcsemath6.test.ts), so the diagrams are accurate, not schematic; students should still be told not to measure from them.
- The proof in section 3 covers the standard 'centre inside the angle' case; strong classes can be shown the other two cases (centre on a chord, centre outside) as the extension of the same isosceles argument.
- Written proof COMPOSITION is classroom work here; the printable worksheets deliberately cover the computable apply-with-reasons side only, because a free-text proof has no computed, never-wrong answer key.