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Maths · Circle Theorems

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Tangents and Chords

Using the tangent and radius, equal tangents, and the perpendicular from the centre to a chord.

  • Higher
  • 9 key terms
  • All boards

Learning Objectives

  1. 1Use the fact that a tangent meets the radius at \(90^\circ\).
  2. 2Use the fact that two tangents from the same point are equal in length.
  3. 3Use the fact that a line from the centre that meets a chord at \(90^\circ\) bisects the chord.
  4. 4Use Pythagoras with radii, tangents and chords.

Tangents and chords

A tangent is a straight line that touches a circle at exactly one point. A chord is a straight line joining two points on a circle. Both have important properties that connect them to the radius. In these questions the radius is usually the key, because it gives you either a right angle or an isosceles triangle. Draw in the radii to the points where the line meets the circle, then look for right-angled triangles and isosceles triangles.

Tangents

A tangent touches the circle at a single point, and is linked to the radius at that point.

  • The theorem

    A tangent to a circle is perpendicular to the radius at the point of contact.

  • Two tangents

    Tangents drawn to a circle from the same point are equal in length.

  • The kite

    The two tangents and two radii make a kite, with two right angles.

  • The reason

    "The angle between a tangent and a radius is \(90^\circ\)." and "Tangents from an external point are equal."

A tangent and Pythagoras

\(PT\) is a tangent to a circle with centre \(O\) and radius 5 cm. \(T\) is the point of contact and \(OP = 13\) cm. Work out the length of \(PT\).

Show the solutionHide the solution
  1. 1 Draw the radius Join \(O\) to \(T\), so \(OT = 5\) cm.
  2. 2 Right angle The tangent meets the radius at \(90^\circ\), so triangle \(OTP\) has a right angle at \(T\).
  3. 3 Hypotenuse \(OP\) is opposite the right angle, so \(OP = 13\) is the hypotenuse.
  4. 4 Pythagoras \(PT^2 = 13^2 - 5^2 = 169 - 25 = 144\), so \(PT = 12\) cm.

Answer\(PT = 12\) cm

Chords

A line from the centre to the middle of a chord is linked to the chord by a right angle.

  • The theorem

    The perpendicular from the centre of a circle to a chord bisects the chord.

  • The reverse

    A line from the centre to the midpoint of a chord is perpendicular to the chord.

  • The triangle

    Joining the centre to both ends of the chord makes an isosceles triangle, because the two sides are radii.

  • The reason

    "The perpendicular from the centre to a chord bisects the chord."

Finding the distance to a chord

A circle has centre \(O\) and radius 10 cm. A chord \(AB\) has length 16 cm. Work out the shortest distance from \(O\) to the chord.

Show the solutionHide the solution
  1. 1 The midpoint The shortest distance \(OM\) meets \(AB\) at a right angle, so \(M\) is the midpoint and \(AM = 8\) cm.
  2. 2 Triangle Triangle \(OMA\) is right-angled, with hypotenuse \(OA = 10\) cm.
  3. 3 Pythagoras \(OM^2 = 10^2 - 8^2 = 100 - 64 = 36\).
  4. 4 Square root \(OM = 6\) cm.

Answer6 cm

Test yourself

  1. 1

    What is the angle between a tangent and the radius?

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    \(90^\circ\).

  2. 2

    What is true of two tangents from the same point?

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    They are equal in length.

  3. 3

    What does a perpendicular from the centre do to a chord?

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    It bisects the chord.

  4. 4

    What shape do two tangents and two radii make?

    Show answerHide answer

    A kite.

  5. 5

    Which theorem do you use to find lengths with these facts?

    Show answerHide answer

    Pythagoras.

Exam technique: lengths and angles

Radii are the key to nearly every tangent and chord question.

  • Draw the radii

    Join the centre to each point of contact, and to the ends of the chord.

  • Mark right angles

    Mark every \(90^\circ\) on the diagram.

  • Pythagoras

    Use it when you need a length in a right-angled triangle.

  • Reasons

    Give the theorem in words for every angle you find.

Summary and exam focus

  • A tangent meets the radius at \(90^\circ\).
  • Tangents from a point are equal in length.
  • The perpendicular from the centre bisects a chord.
  • Use Pythagoras in the right-angled triangles that result.

Exam focus

\(PA\) and \(PB\) are tangents to a circle with centre \(O\). Angle \(AOB = 120^\circ\). Work out angle \(APB\), and give reasons. (3 marks) (3 marks)

The angles at \(A\) and \(B\) are \(90^\circ\), because a tangent meets the radius at \(90^\circ\). Then \(APB = 360 - 90 - 90 - 120 = 60^\circ\), because the angles in a quadrilateral add up to \(360^\circ\).

Key terms

The words this lesson expects you to use. Each one is linked from the first place it appears above.

Tangent
A straight line that touches a circle at one point.
Chord
A straight line joining two points on a circle.
Point of contact
The point where a tangent touches the circle.
Perpendicular
At \(90^\circ\) to a line.
Bisect
Cut exactly in half.
Radius
The distance from the centre to the circle.
Kite
A quadrilateral with two pairs of equal adjacent sides.
Hypotenuse
The longest side of a right-angled triangle.
Isosceles triangle
A triangle with two equal sides.

Questions and answers

15 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Work out 3 marks Core

Diagram NOT accurately drawn. \(TA\) and \(TB\) are tangents to a circle, centre \(O\). Angle \(ATB = 64^\circ\). Work out the size of angle \(TAB\). Give a reason for your answer. (3 marks)

A circle diagram showing two tangents from a point T.

Mark scheme — 3 marks available

  • Tangents from a point are equal, so triangle TAB is isosceles — C1
  • \((180 - 64) \div 2\) — M1
  • \(58\) — A1

Model answer

\(TA = TB\) because tangents from a point to a circle are equal, so triangle \(TAB\) is isosceles. Angle \(TAB = (180 - 64) \div 2 = 58^\circ\).

2. Exam question Work out 3 marks Core

\(PT\) is a tangent to a circle, centre \(O\), at the point \(T\). The radius of the circle is 6 cm and \(OP = 10\) cm. Work out the length of \(PT\). (3 marks)

Mark scheme — 3 marks available

  • Angle OTP is a right angle, because a tangent is perpendicular to the radius — C1
  • \(10^2 - 6^2\) or \(100 - 36\) — M1
  • \(8\) — A1

Model answer

The tangent is perpendicular to the radius, so angle \(OTP = 90^\circ\). \(PT^2 = 10^2 - 6^2 = 64\), so \(PT = 8\) cm.

3. Exam question Work out 3 marks Core

Diagram NOT accurately drawn. \(AB\) is a chord of a circle, centre \(O\), with radius 13 cm. \(AB = 24\) cm. \(M\) is the point on \(AB\) where \(OM\) is perpendicular to \(AB\). Work out the length of \(OM\). (3 marks)

A circle diagram showing a chord and the perpendicular from the centre.

Mark scheme — 3 marks available

  • \(AM = 12\) — M1
  • \(13^2 - 12^2\) — M1
  • \(5\) — A1

Model answer

The perpendicular from the centre bisects the chord, so \(AM = 12\) cm. \(OM^2 = 13^2 - 12^2 = 169 - 144 = 25\), so \(OM = 5\) cm.

4. Exam question Work out 4 marks Core

Diagram NOT accurately drawn. \(PA\) and \(PB\) are tangents to a circle, centre \(O\). Angle \(OAB = 35^\circ\). Work out the size of angle \(APB\). Give reasons for your answer. (4 marks)

A circle diagram showing two tangents and a chord.

Mark scheme — 4 marks available

  • Angle \(OAP = 90^\circ\), as a tangent is perpendicular to the radius — C1
  • \(PAB = 90 - 35 = 55\) — M1
  • \(180 - 55 - 55\) — M1
  • \(70\) — A1

Model answer

Angle \(OAP = 90^\circ\), because a tangent is perpendicular to the radius. So \(PAB = 90 - 35 = 55^\circ\). \(PA = PB\) because tangents from a point are equal, so \(PBA = 55^\circ\) and \(APB = 180 - 55 - 55 = 70^\circ\).

5. Exam question Work out 3 marks Stretch

A circle has centre \(O\) and radius 25 cm. A chord is 7 cm from \(O\). Work out the length of the chord. (3 marks)

Mark scheme — 3 marks available

  • \(25^2 - 7^2\) or \(625 - 49\) — M1
  • \(24\) found as half the chord — M1
  • \(48\) — A1

Model answer

Half the chord is \(\sqrt{25^2 - 7^2} = \sqrt{576} = 24\) cm, so the chord is \(2 \times 24 = 48\) cm.

6. Exam question Prove 4 marks Stretch

\(TA\) and \(TB\) are tangents to a circle, centre \(O\), touching the circle at \(A\) and \(B\). Prove that triangles \(OAT\) and \(OBT\) are congruent, and hence that \(TA = TB\). (4 marks)

Mark scheme — 4 marks available

  • OA = OB because they are radii — B1
  • Angles \(OAT = OBT = 90^\circ\), because a tangent is perpendicular to the radius — B1
  • OT is common, so the triangles are congruent (right angle, hypotenuse, side) — B1
  • Concludes TA = TB because corresponding sides of congruent triangles are equal — C1

Model answer

\(OA = OB\), because they are radii. Angles \(OAT\) and \(OBT\) are both \(90^\circ\), because a tangent is perpendicular to the radius. \(OT\) is common to both triangles. So the triangles are congruent (RHS), and \(TA = TB\).

7. Multiple choice 1 mark Easier

What is the angle between a tangent and the radius at the point of contact?

  1. A \(45^\circ\)
  2. B \(180^\circ\)
  3. C \(60^\circ\)
  4. D \(90^\circ\) Correct

Why: A tangent is perpendicular to the radius.

8. Multiple choice 1 mark Easier

Two tangents from the point \(P\) touch a circle at \(A\) and \(B\). \(PA = 9\) cm. What is \(PB\)?

  1. A 18 cm
  2. B 4.5 cm
  3. C 9 cm Correct
  4. D It cannot be found

Why: Tangents from the same point are equal in length.

9. Multiple choice 1 mark Easier

A perpendicular from the centre of a circle meets a chord. What does it do to the chord?

  1. A It doubles the chord
  2. B It bisects the chord Correct
  3. C It is the same length as the chord
  4. D It makes the chord a diameter

Why: The perpendicular from the centre bisects the chord.

10. Multiple choice 1 mark Core

\(PT\) is a tangent, \(O\) is the centre, the radius is 3 cm and \(OP = 5\) cm. How long is \(PT\)?

  1. A 4 cm Correct
  2. B 2 cm
  3. C \(\sqrt{34}\) cm
  4. D 8 cm

Why: \(OTP\) is right-angled at \(T\), so \(PT^2 = 5^2 - 3^2 = 16\).

11. Multiple choice 1 mark Core

Tangents \(PA\) and \(PB\) touch a circle with centre \(O\). Angle \(AOB = 100^\circ\). What is angle \(APB\)?

  1. A \(100^\circ\)
  2. B \(50^\circ\)
  3. C \(260^\circ\)
  4. D \(80^\circ\) Correct

Why: \(OAPB\) is a quadrilateral with two right angles, so \(APB = 360 - 90 - 90 - 100 = 80^\circ\).

12. Multiple choice 1 mark Core

A circle has radius 5 cm and a chord is 8 cm long. How far is the chord from the centre?

  1. A 4 cm
  2. B \(\sqrt{41}\) cm
  3. C 3 cm Correct
  4. D 1 cm

Why: Half the chord is 4 cm, so the distance is \(\sqrt{5^2 - 4^2} = 3\).

13. Multiple choice 1 mark Core

Tangents \(PA\) and \(PB\) touch a circle at \(A\) and \(B\). Angle \(APB = 50^\circ\). What is angle \(PAB\)?

  1. A \(50^\circ\)
  2. B \(65^\circ\) Correct
  3. C \(130^\circ\)
  4. D \(40^\circ\)

Why: \(PA = PB\), so triangle \(PAB\) is isosceles and \(PAB = (180 - 50) \div 2 = 65^\circ\).

14. Multiple choice 1 mark Stretch

The distance from the centre of a circle of radius 13 cm to a chord is 5 cm. How long is the chord?

  1. A 24 cm Correct
  2. B 12 cm
  3. C \(2\sqrt{194}\) cm
  4. D 10 cm

Why: Half the chord is \(\sqrt{13^2 - 5^2} = 12\), so the chord is 24 cm.

15. Multiple choice 1 mark Stretch

A line from the centre to a point \(P\) outside a circle of radius \(r\) has length \(d\). Which expression gives the tangent length from \(P\)?

  1. A \(\sqrt{d^2 + r^2}\)
  2. B \(d - r\)
  3. C \(d + r\)
  4. D \(\sqrt{d^2 - r^2}\) Correct

Why: The radius and tangent make a right angle, with \(d\) as the hypotenuse.