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Cambridge IGCSE0580

Circle theorems

Mathematics 0580 Chapter Notes

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Circle theorems
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Circle theorems notes

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In this chapter

  1. 1Angle in a semicircle
  2. 2Tangent and radius
  3. 3Angle at the centre
  4. 4Angles in the same segment
  5. 5Cyclic quadrilateral
  6. 6Alternate segment theorem

1. Angle in a semicircle

The angle in a semicircle is 90°: join the two ends of a diameter to any point on the circumference and the angle at that point is a right angle.

Always say angle, not triangle, and mention the 90° (or right angle) in your reason.

Once you have a right angle, the angles in a triangle add to 180° gives the rest.

angle in a semicircle = 90°

Examiner insight

“The reason was often not expressed accurately and often a triangle, rather than an angle, in a semicircle was given. The explanation also needed reference to the angle of 90° (or right angle).”

Worked example 13 marks

A, B, C and D are points on the circumference of a circle with centre O. AB = AD and angle BDC = 52°. BD is a diameter of the circle. Calculate angle ABC.

  1. 1

    Angle BCD = 90° (angle in a semicircle), so angle DBC = 180 - 90 - 52 = 38° [1]

  2. 2

    Angle BAD = 90° (angle in a semicircle) and AB = AD, so angle ABD = (180 - 90) ÷ 2 = 45° [1]

  3. 3

    Angle ABC = 45 + 38 = 83° [1]

2. Tangent and radius

A tangent touches the circle at one point. The angle between tangent and radius = 90° at that point.

When you give the reason, name both the tangent and the radius. Writing only "tangent" earns nothing. Then use angles in a triangle add to 180° to find the other angles.

angle between tangent and radius = 90°

Examiner insight

“Again the majority of candidates were able to state that angle SQP = 27° but attempts to give the reason were often insufficient, frequently mentioning only tangent.”

Worked example 13 marks

PT is a tangent to a circle, centre O, touching the circle at T. Angle TOP = 63°. Find angle TPO. Give a reason for each step.

  1. 1

    Angle OTP = 90° (angle between tangent and radius = 90°) [1]

  2. 2

    Angle TPO = 180 - 90 - 63 (angles in a triangle add to 180°) [1]

  3. 3

    Angle TPO = 27° [1]

3. Angle at the centre

The angle at the centre is twice the angle at the circumference, when both stand on the same arc.

Work from the arc the angles stand on. If the point on the circumference lies on the minor arc, the angle at the centre is the reflex angle. A reflex angle is always more than 180°, so check that your answer makes sense.

angle at centre = 2 × angle at circumference

Examiner insight

“many did not recognise that this was an example of the angle at the centre and that angle ACB was the corresponding angle at the circumference.” “Although the diagram is not to scale candidates are advised to check the common sense of their answer; a clearly reflex angle would not be less than 180°.”

Worked example 12 marks

A, B and C are points on a circle, centre O. C is on the major arc AB. Angle AOB = 124°. Find angle ACB. Give a reason.

  1. 1

    Angle at the centre is twice the angle at the circumference [1]

  2. 2

    Angle ACB = 124 ÷ 2 = 62° [1]

4. Angles in the same segment

Angles in the same segment are equal. The angles that a chord makes at two points on the circumference, both on the same side of the chord, are equal.

Check that both points lie on the same side of the chord before you use this. Angles on opposite sides of the chord are opposite angles of a cyclic quadrilateral, so they sum to 180° instead.

∠ ACB = ∠ ADB

Examiner insight

“Another common incorrect method was to think that OAP and BCP were angles in the same segment and so w = 21° was also a common incorrect answer.”

Worked example 12 marks

A, B, C and D are points on a circle. C and D are on the same side of the chord AB. Angle ACB = 38°. Find angle ADB. Give a reason.

  1. 1

    Angles in the same segment are equal [1]

  2. 2

    Angle ADB = 38° [1]

5. Cyclic quadrilateral

A cyclic quadrilateral has all four corners on a circle. Opposite angles of a cyclic quadrilateral sum to 180° (they are supplementary).

Say cyclic and say angles. It is the opposite angles that sum to 180°, not the opposite sides.

∠ A + ∠ C = 180° ∠ B + ∠ D = 180°

Examiner insight

“Some candidates described opposite sides, rather than opposite angles, as summing to 180°.” “A small number omitted the word ‘cyclic’ when describing the quadrilateral.”

Worked example 12 marks

ABCD is a cyclic quadrilateral. Angle ABC = 112° and angle BCD = 95°. Find angle ADC and angle BAD. Give a reason.

  1. 1

    Opposite angles of a cyclic quadrilateral sum to 180°: angle ADC = 180 - 112 = 68° [1]

  2. 2

    Angle BAD = 180 - 95 = 85° [1]

6. Alternate segment theorem

The angle between a tangent and a chord is equal to the angle in the alternate segment. The alternate segment is the one on the other side of the chord from the angle.

Name the theorem in full: alternate segment theorem. It is not the alternate angle theorem. Alternate angles belong to parallel lines.

tangent–chord angle = angle in the alternate segment

Examiner insight

“Some candidates provided a fully correct explanation for their answer of 116°, but it was more common to see a partially correct explanation, with many omitting to mention the alternate segment theorem or incorrectly calling it the alternate angle theorem.” “It is important, in a question such as this, that candidates include the names of the theorems being used when giving reasons for their answer;”

Worked example 13 marks

TA is a tangent to a circle at A. B and C are points on the circle, with C on the opposite side of the chord AB from T. Angle TAB = 48° and angle ABC = 75°. Find angle BAC. Give a reason for each step.

  1. 1

    Angle ACB = 48° (alternate segment theorem) [1]

  2. 2

    Angle BAC = 180 - 48 - 75 (angles in a triangle add to 180°) [1]

  3. 3

    Angle BAC = 57° [1]

Recap

  • Angle in a semicircle = 90°. Say angle, not triangle.
  • Angle between tangent and radius = 90°.
  • Extended: angle at the centre is twice the angle at the circumference.
  • Extended: angles in the same segment are equal.
  • Extended: opposite angles of a cyclic quadrilateral sum to 180°.
  • Extended: alternate segment theorem. Name the theorem in your reason.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Do not use a calculator. State the size of the angle between a tangent to a circle and the radius drawn to the point of contact.1 mark
  2. Do not use a calculator. AB is a diameter of a circle and C is a point on the circumference. Angle CAB = 37°. Find angle ABC.2 marks
  3. PT is a tangent to a circle, centre O, touching it at T. Angle OPT = 35°. Find angle TOP. Give a reason for each step.3 marks
  4. Do not use a calculator. AB is a diameter of a circle and C is a point on the circumference. AC = BC. Find angle CAB.3 marks
  5. Do not use a calculator. PAQ is a tangent to a circle, centre O, at A. C is a point on the circle on the same side of OA as P. Angle OAC = 20°. Find angle PAC.3 marks
  6. Do not use a calculator. A, B and C are points on a circle, centre O. C is on the major arc AB. Angle ACB = 71°. Find angle AOB.2 marks
  7. A, B and C are points on a circle, centre O. C is on the minor arc AB. Angle ACB = 118°. Find the reflex angle AOB.2 marks
  8. Do not use a calculator. A, B, C and D are points on a circle. C and D are on the same side of the chord AB. Angle ADB = (2x + 10)° and angle ACB = (3x - 15)°. Find x and angle ACB.3 marks
  9. A, B, C and D are points on a circle, centre O, in that order round the circle. Angle ABC = 124°. Find angle OAC. Give a reason for each step.4 marks
  10. A, B, C and D are points on a circle, in that order round the circle. PAQ is the tangent at A, with P on the opposite side of AB from D. Angle PAB = (x + 10)°, angle QAD = (2x + 5)° and angle BCD = (4x - 15)°. Find x and angle ABD. Give a reason for each step.5 marks

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