Circle angle facts let you find an angle from a single clue, such as a diameter or a centre. Each fact has its own wording, and a correct reason names that fact. Common ones are: the angle at the centre is twice the angle at the circumference; the angle in a semicircle is 90°; opposite angles of a cyclic quadrilateral add to 180°; and a tangent is perpendicular to the radius at the point of contact.
This is the fourth lesson in angles and geometric reasoning. Check the Cambridge syllabus page for which of these belong to your Core or Extended route.
Which circle facts should you know?
| Clue in the diagram | Fact | Reason to write |
|---|---|---|
| Centre O, same arc | Angle at centre = 2 × angle at circumference | The angle at the centre is twice the angle at the circumference |
| Diameter | Angle is 90° | The angle in a semicircle is 90° |
| Four points on the circle | Opposite angles add to 180° | Opposite angles of a cyclic quadrilateral add to 180° |
| Tangent | Meets radius at 90° | A tangent is perpendicular to the radius |
How do you choose the right fact?
- Mark any centre, diameter or tangent in the diagram.
- Find the arc or chord your known angle stands on.
- Match the clue to a fact in the table.
- Write the calculation and the reason together.
- Check the size: an angle at the centre should be larger than the corresponding angle at the circumference.
Worked example
O is the centre of a circle. A, B and C lie on the circle, with C on the major arc.
∠AOB = 100°. A fourth point D lies on the minor arc AB. Find ∠ACB and ∠ADB, with reasons.
Step 1: ∠ACB and ∠AOB both stand on arc AB.
Step 2: The angle at the centre is twice the angle at the circumference, so ∠ACB = 100° ÷ 2 = 50°.
Step 3: ACBD is a cyclic quadrilateral, and C and D are opposite corners.
Step 4: Opposite angles of a cyclic quadrilateral add to 180°, so ∠ADB = 180° − 50° = 130°.
Check: ∠ADB is on the minor arc, so it is obtuse, and 130° fits a wide angle seen from the short side.
The mistake to watch for
A common slip is to double the angle when you should halve it.
Mistaken answer: ∠ACB = 2 × 100° = 200°.
The student knew the word “twice” but used it backwards. The angle at the centre is the larger one.
The correction is to ask which angle is at the centre. That angle is the double, so the circumference angle is half.
A quick sense check also helps: 200° is more than a straight line, which is not possible for an angle inside a triangle. A reason like “circle theorem” alone also loses the mark, so write the full fact.
Check yourself
Try these, then open each answer.
1. AB is a diameter of a circle, and C is a point on the circle. ∠CAB = 35°. Find ∠ABC, with a reason.
Show answer
∠ACB = 90°, because the angle in a semicircle is 90°. Then ∠ABC = 180° − 90° − 35° = 55°, because the angles in a triangle add to 180°.
2. ABCD is a cyclic quadrilateral with ∠ABC = 78°. Find ∠ADC.
Show answer
B and D are opposite corners, and opposite angles of a cyclic quadrilateral add to 180°. So ∠ADC = 180° − 78° = 102°.
3. TA is a tangent to a circle with centre O, touching at A. ∠ATO = 32°. Find ∠AOT.
Show answer
A tangent is perpendicular to the radius at A, so ∠OAT = 90°. Then ∠AOT = 180° − 90° − 32° = 58°, using the angle sum of triangle OAT.
Where this leads next
The last lesson in the module asks you to stay honest about what a diagram shows: separate a diagram assumption from a stated fact. The non-calculator working trainer supports the halving and subtracting, and the mixed practice set includes circle questions beside the rest.
Some students can name every theorem but struggle to pick one from a crowded diagram. Our teachers practise that choice in online one-to-one Mathematics tuition.