📘 GCSE Maths · Notes & worksheet
Circle Theorems.
Notes
The theorems
- Angle at the centre is twice the angle at the circumference, when both stand on the same arc.
- Angles in the same segment are equal (they stand on the same chord).
- Angle in a semicircle is 90°. The diameter subtends a right angle at the circumference.
- Cyclic quadrilateral: opposite angles add up to 180°.
- Tangent and radius meet at 90°.
- Tangents from an external point are equal in length.
- Alternate segment theorem: the angle between a tangent and a chord equals the angle in the alternate segment.
- Perpendicular from the centre to a chord bisects the chord.
How to write a reason
Examiners need the theorem name or statement for every step. "Angle at centre is twice angle at circumference" scores; "because it's a circle" does not.
Strategy
- Mark every radius: radii are equal, so they create isosceles triangles.
- Look for right angles from tangents and diameters.
- Work step by step, writing the reason beside each angle you find.
Worked example
OAPB is a kite where PA and PB are tangents from P to a circle centre O, and angle APB = 50°.
- Angles OAP and OBP are 90° (tangent ⟂ radius).
- Angles in a quadrilateral add to 360°, so angle AOB = 360 − 90 − 90 − 50 = 130°.
Proof template: angle in a semicircle is 90°
Let AB be a diameter, centre O, and C on the circumference. OA = OB = OC (radii), so triangles OAC and OBC are isosceles. Let angle OAC = x, giving angle OCA = x, and let angle OBC = y, giving angle OCB = y. In triangle ABC the angles sum to 180°: x + y + (x + y) = 180, so x + y = 90 and angle ACB = 90°.
Common mistakes
- Using the wrong arc for the "angle at the centre" theorem.
- Quoting two reasons in one line without saying which angle each applies to.
- Forgetting the alternate segment theorem uses the angle on the opposite side of the chord.
Worksheet
Prints the questions only, without answers. Choose “Save as PDF” in the print window to download.
Diagram placeholderThe diagrams for all questions are described in words in the questions. The drawn diagrams are being added to this page.
Total: 21 marks. Give a reason for every angle you find.
- A, B and C lie on a circle with centre O. C is on the major arc AB. Angle AOB = 112°. Find angle ACB. (2)
- PQRS is a cyclic quadrilateral. Angle PQR = 105° and angle QRS = 88°. Find angle PSR and angle QPS. (3)
- AB is a diameter of a circle and C is on the circumference. Angle CAB = 34°. Find angle ABC. (2)
- TA is a tangent to a circle at A, and AB is a chord. Angle TAB = 65°. C is a point on the circle in the alternate segment, and O is the centre.
(a) Find angle ACB. (2)
(b) Find angle AOB. (1) - Tangents from an external point P touch a circle, centre O, at A and B. Angle APB = 50°. Find angle AOB, angle OAB and angle PAB. (4)
- A circle has radius 13 cm. A chord AB has length 24 cm. Find the perpendicular distance from the centre to the chord. (3)
- Prove that the angle in a semicircle is 90°. (4)
Show mark scheme and answers (tutor copy)
Try the worksheet first, then check your work here.
- 56°. Angle at the centre is twice the angle at the circumference. (M1 theorem, A1)
- Opposite angles of a cyclic quadrilateral sum to 180°. PSR = 180 − 105 = 75°. (M1, A1) QPS = 180 − 88 = 92°. (A1)
- Angle ACB = 90° (angle in a semicircle) (M1). ABC = 180 − 90 − 34 = 56°. (A1)
- (a) 65° by the alternate segment theorem. (M1, A1)
(b) 130°, since the angle at the centre is twice the angle at the circumference. (B1) - Angles OAP and OBP are 90° (tangent ⟂ radius). AOB = 360 − 90 − 90 − 50 = 130°. (M1, A1) Triangle OAB is isosceles (OA = OB), so OAB = (180 − 130)/2 = 25°. (M1) PAB = 90 − 25 = 65°. (A1) Check: PA = PB, so PAB = (180 − 50)/2 = 65° ✓.
- The perpendicular from the centre bisects the chord, so the half-chord is 12 cm (M1). Distance² = 13² − 12² = 169 − 144 = 25 (M1). Distance = 5 cm. (A1)
- See the proof template in the notes. Marks: B1 OA = OB = OC (radii) so the triangles are isosceles. B1 base angles equal, naming x and y. B1 angle sum of triangle ABC gives 2x + 2y = 180. B1 conclude x + y = 90, so angle ACB = 90°.