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Exam questions · Maths · Circle Theorems

Tangents and Chords

  • 6 exam questions
  • 20 marks
  • 9 quick checks
  1. 1 Work out [3 marks]

    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.
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    Model answer

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

    Mark scheme

    • Tangents from a point are equal, so triangle TAB is isosceles — C1
    • \((180 - 64) \div 2\) — M1
    • \(58\) — A1
  2. 2 Work out [3 marks]

    \(PT\) 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)

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    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.

    Mark scheme

    • 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
  3. 3 Work out [3 marks]

    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.
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    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.

    Mark scheme

    • \(AM = 12\) — M1
    • \(13^2 - 12^2\) — M1
    • \(5\) — A1
  4. 4 Work out [4 marks]

    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.
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    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 \(APB = 180 - 55 - 55 = 70^\circ\).

    Mark scheme

    • Angle \(OAP = 90^\circ\), as a tangent is perpendicular to the radius — C1
    • \(PAB = 90 - 35 = 55\) — M1
    • \(180 - 55 - 55\) — M1
    • \(70\) — A1
  5. 5 Work out [3 marks]

    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)

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    Model answer

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

    Mark scheme

    • \(25^2 - 7^2\) or \(625 - 49\) — M1
    • \(24\) found as half the chord — M1
    • \(48\) — A1
  6. 6 Prove [4 marks]

    \(TA\) 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)

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    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 \(TA = TB\).

    Mark scheme

    • 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

Quick check

  1. 1

    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\(90^\circ\)
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    D: \(90^\circ\)

    A tangent is perpendicular to the radius.

  2. 2

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

    1. A18 cm
    2. B4.5 cm
    3. C9 cm
    4. DIt cannot be found
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    C: 9 cm

    Tangents from the same point are equal in length.

  3. 3

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

    1. AIt doubles the chord
    2. BIt bisects the chord
    3. CIt is the same length as the chord
    4. DIt makes the chord a diameter
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    B: It bisects the chord

    The perpendicular from the centre bisects the chord.

  4. 4

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

    1. A4 cm
    2. B2 cm
    3. C\(\sqrt{34}\) cm
    4. D8 cm
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    A: 4 cm

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

  5. 5

    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\(80^\circ\)
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    D: \(80^\circ\)

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

  6. 6

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

    1. A4 cm
    2. B\(\sqrt{41}\) cm
    3. C3 cm
    4. D1 cm
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    C: 3 cm

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

  7. 7

    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\(65^\circ\)
    3. C\(130^\circ\)
    4. D\(40^\circ\)
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    B: \(65^\circ\)

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

  8. 8

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

    1. A24 cm
    2. B12 cm
    3. C\(2\sqrt{194}\) cm
    4. D10 cm
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    A: 24 cm

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

  9. 9

    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\(\sqrt{d^2 - r^2}\)
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    D: \(\sqrt{d^2 - r^2}\)

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