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Cambridge IGCSE0607

Circle theorems II

International Mathematics 0607 Chapter Notes

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Circle theorems II
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Circle theorems II notes

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In this chapter

  1. 1Tangents from an external point
  2. 2Using equal tangents to find a length
  3. 3The perpendicular bisector of a chord
  4. 4Equal chords are equidistant from the centre

1. Tangents from an external point

A tangent touches a circle at one point. From a point outside the circle you can draw two tangents.

Tangents from an external point are equal in length. Each tangent also meets the radius at the point of contact at a right angle: angle between tangent and radius = 90°.

In a question, write the reason in words each time you use the fact. A calculation alone does not earn the mark.

BA = BC

Examiner insight

“Candidates must use the correct language and be precise.” “Words and phrases such as perimeter, edge, top of the circle, middle, origin or triangle were not accepted for circumference, centre or angle.”

Worked example 17 marks

A circle has centre O and radius 6.4 cm. B is a point outside the circle. BA and BC are tangents to the circle, with A and C on the circle. AB is 11.3 cm.

(a) Write down the length of CB.

(b) Give a geometrical reason why angle OAB = 90°.

(c) Work out the size of angle ABC.

(d) Work out the length of OB.

  1. 1

    (a) CB = 11.3 cm (tangents from an external point are equal in length) [1]

  2. 2

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

  3. 3

    (c) triangle OAB has a right angle at A, so tan(∠ OBA) = 6.4/11.3 [1]

  4. 4

    enter tan⁻¹(6.4 ÷ 11.3) on the calculator and read ∠ OBA = 29.526...^(°) [1]

  5. 5

    OB bisects angle ABC, so ∠ ABC = 2 × 29.526... = 59.1^(°) [1]

  6. 6

    (d) OB² = 6.4² + 11.3² = 168.65 [1]

  7. 7

    OB = √(168.65) = 13.0 cm [1]

2. Using equal tangents to find a length

When a circle touches the sides of a shape, each corner has two equal tangent lengths. Mark them with the same letter or expression.

A side of the shape is then the sum of two tangent lengths. This gives an equation for the unknown.

Give the reason as tangents from an external point are equal in length.

Examiner insight

“Many candidates did not use correct angle notation and simply referred to angles as e.g. D.” “In addition, candidates are expected to give correct mathematical reasons for angles being equal in any geometrical proof.”

Worked example 15 marks

Do not use a calculator. ABC is a right-angled triangle with the right angle at A. AB = 6 cm, AC = 8 cm and BC = 10 cm. A circle, centre O and radius r cm, touches AB at D, BC at E and AC at F.

(a) Give a geometrical reason to explain why BD = BE.

(b) Find the area of the circle. Give your answer in terms of π.

  1. 1

    (a) tangents from an external point are equal in length [1]

  2. 2

    (b) OD and OF are radii at right angles to the sides, and angle A = 90^(°), so ADOF is a square and AD = AF = r [1]

  3. 3

    BE = BD = 6 - r and CE = CF = 8 - r [1]

  4. 4

    BC = BE + EC, so (6 -r) + (8 -r) = 10 [1]

  5. 5

    14 - 2r = 10, so r = 2 and area = π × 2² = 4π cm² [1]

3. The perpendicular bisector of a chord

A chord is a straight line joining two points on a circle. The perpendicular bisector of a chord passes through the centre.

So a line from the centre at right angles to a chord cuts the chord in half. Joining the centre to the end of the chord gives a right-angled triangle: the radius is the hypotenuse.

Two chords give two bisectors. They meet at the centre.

Examiner insight

“The most common errors came from slips with arithmetic, signs and cancelling, finding the equation of the line AB, using point A or B instead of the midpoint and errors in finding the midpoint.”

Worked example 13 marks

Do not use a calculator. A chord AB of a circle, centre O, is 24 cm long. The radius of the circle is 13 cm. Find the perpendicular distance from O to AB.

  1. 1

    the perpendicular from O meets AB at its midpoint M, so AM = 24 ÷ 2 = 12 [1]

  2. 2

    triangle OMA has a right angle at M: OM² = 13² - 12² = 169 - 144 = 25 [1]

  3. 3

    OM = 5 cm [1]

4. Equal chords are equidistant from the centre

Equal chords are equidistant from the centre. The distance is measured along the perpendicular from the centre to the chord.

So when two chords are equal and you know how far one is from the centre, you know the distance for the other. Write the property as your reason.

To find the distance, use the right-angled triangle made by the half-chord, the distance and the radius.

Worked example 14 marks

Do not use a calculator. Two chords PQ and RS of a circle, centre O, are equal in length. The radius is 25 cm and PQ = 48 cm. Find the perpendicular distance from O to RS. Give a geometrical reason for your answer.

  1. 1

    the perpendicular from O cuts PQ in half: 48 ÷ 2 = 24 [1]

  2. 2

    distance from O to PQ: d² = 25² - 24² = 625 - 576 = 49 [1]

  3. 3

    d = 7, so PQ is 7 cm from O [1]

  4. 4

    equal chords are equidistant from the centre, so RS is also 7 cm from O [1]

Recap

  • Tangents from an external point are equal in length.
  • The angle between a tangent and the radius at the point of contact is 90°.
  • Always write the reason in words. A calculation alone is not enough.
  • Extended: the perpendicular bisector of a chord passes through the centre.
  • Extended: the half-chord, the distance from the centre and the radius make a right-angled triangle.
  • Extended: equal chords are equidistant from the centre.

End-of-chapter exercise

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

  1. Complete the sentence. Tangents from an external point are ...1 mark
  2. Two tangents are drawn from a point P to a circle. They touch the circle at A and B. PA = 9.5 cm. Write down the length of PB and give a reason.2 marks
  3. Do not use a calculator. Two tangents from a point T touch a circle, centre O, at P and Q. The radius OP is 5 cm and PT is 12 cm. Find OT.3 marks
  4. Two tangents from a point X touch a circle at Y and Z. XY = (3x + 2) cm and XZ = (5x - 10) cm. Find x and the length of XY.3 marks
  5. Do not use a calculator. A circle touches the sides of triangle ABC at D on AB, E on BC and F on CA. AD = 4 cm, BE = 5 cm and CF = 7 cm. Find the perimeter of the triangle.3 marks
  6. Chords AB and CD of a circle, centre O, are equal in length. The perpendicular distance from O to AB is 9 cm. Write down the distance from O to CD and give a reason.2 marks
  7. Do not use a calculator. A chord of a circle is 16 cm long. The radius is 10 cm. Find the perpendicular distance from the centre to the chord.3 marks
  8. Do not use a calculator. A, B and C are points on a circle. A is (0, 0), B is (6, 0) and C is (6, 8). Find the coordinates of the centre and the radius.4 marks
  9. Do not use a calculator. The arch of a bridge is an arc of a circle, centre O. At water level the arch is 24 m wide: this width is a chord AB of the circle. The highest point H of the arch is 8 m above M, the midpoint of AB. Find the radius of the circle. Give a geometrical reason for the position of O.5 marks
  10. Do not use a calculator. A ball rests in a cone-shaped holder. In the cross-section the ball is a circle, centre O, radius 15 cm. The sides of the holder are tangents TP and TQ from the vertex T, touching the circle at P and Q. TP = 20 cm. OT crosses the chord PQ at M. (a) Find OT. (b) Explain why OT is the perpendicular bisector of PQ. Find the length of PQ. (c) A chord RS of the circle is equal in length to PQ. Find the distance from O to RS. Give a geometrical reason.7 marks

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