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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.

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  • 9 key terms
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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 \(PT\).

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  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.

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  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?

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    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.

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Test yourself while it is fresh. Start with the flashcards, then try the exam questions.

Flashcards 16 cards to flip through Exam questions 6 questions, 20 marks with answers
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