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Exam questions · Maths · Circle Theorems

Circle Theorem Proofs and Problems

  • 6 exam questions
  • 24 marks
  • 9 quick checks
  1. 1 Prove [4 marks]

    Diagram NOT accurately drawn. \(A\), \(B\) and \(C\) are points on a circle, centre \(O\). Prove that angle \(AOB\) is twice angle \(ACB\). You may add lines to the diagram. (4 marks)

    A circle diagram showing the angle at the centre and the angle at the circumference, for a proof.
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    Model answer

    Draw the line \(CO\) and extend it to meet the circle at \(D\). \(OA = OC = OB\), because they are radii. Triangles \(OAC\) and \(OBC\) are isosceles, so \(OAC = OCA\) and \(OBC = OCB\). The exterior angle \(AOD = 2 \times OCA\) and \(BOD = 2 \times OCB\). So \(AOB = 2(OCA + OCB) = 2 \times ACB\).

    Mark scheme

    • Draws CO extended to D and states OA = OB = OC as radii — B1
    • Isosceles triangles, so OCA = OAC and OCB = OBC — M1
    • Exterior angles: AOD = 2 x OCA and BOD = 2 x OCB — M1
    • Concludes AOB = 2 x ACB — C1
  2. 2 Work out [4 marks]

    Diagram NOT accurately drawn. \(A\), \(B\), \(C\) and \(D\) are points on a circle, centre \(O\). Angle \(AOC = 148^\circ\). (a) Work out the size of angle \(ABC\). Give a reason for your answer. (2 marks) (b) Work out the size of angle \(ADC\). Give a reason for your answer. (2 marks)

    A circle diagram showing a cyclic quadrilateral and its centre.
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    Model answer

    (a) \(ABC = 148 \div 2 = 74^\circ\), because the angle at the centre is twice the angle at the circumference. (b) \(ADC = 180 - 74 = 106^\circ\), because opposite angles of a cyclic quadrilateral add up to \(180^\circ\).

    Mark scheme

    • (a) \(74\) — B1
    • (a) The angle at the centre is twice the angle at the circumference — C1
    • (b) \(106\) — B1
    • (b) Opposite angles of a cyclic quadrilateral add up to 180 degrees — C1
  3. 3 Find [4 marks]

    The point \(P(4, 3)\) is on the circle \(x^2 + y^2 = 25\). Find an equation of the tangent to the circle at \(P\). (4 marks)

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    Model answer

    The radius \(OP\) has gradient \(\dfrac{3}{4}\), so the tangent has gradient \(-\dfrac{4}{3}\). Then \(y - 3 = -\dfrac{4}{3}(x - 4)\), which gives \(y = -\dfrac{4}{3}x + \dfrac{25}{3}\).

    Mark scheme

    • Gradient of \(OP = \dfrac{3}{4}\) — B1
    • Gradient of the tangent \(= -\dfrac{4}{3}\) — B1
    • \(y - 3 = -\dfrac{4}{3}(x - 4)\) or equivalent — M1
    • \(y = -\dfrac{4}{3}x + \dfrac{25}{3}\) or \(4x + 3y = 25\) — A1
  4. 4 Show that [3 marks]

    Show that the line \(y = -\dfrac{3}{4}x + \dfrac{25}{2}\) is a tangent to the circle \(x^2 + y^2 = 100\) at the point \((6, 8)\). (3 marks)

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    Model answer

    \(6^2 + 8^2 = 100\), so \((6, 8)\) is on the circle. On the line, \(-\dfrac{3}{4} \times 6 + \dfrac{25}{2} = -4.5 + 12.5 = 8\), so the point is on the line. The radius has gradient \(\dfrac{8}{6} = \dfrac{4}{3}\) and \(\dfrac{4}{3} \times -\dfrac{3}{4} = -1\), so the line is perpendicular to the radius, and so is a tangent.

    Mark scheme

    • Checks that \((6, 8)\) is on both the circle and the line — B1
    • Gradient of radius \(= \dfrac{4}{3}\) — B1
    • Product of the gradients is \(-1\), so the line is perpendicular to the radius, with a conclusion — C1
  5. 5 Work out [5 marks]

    Diagram NOT accurately drawn. \(TAS\) is a tangent to the circle at \(A\). \(A\), \(B\), \(C\) and \(D\) are points on the circle. Angle \(TAD = 52^\circ\) and angle \(CAD = 43^\circ\). (a) Write down the size of angle \(ACD\). Give a reason for your answer. (2 marks) (b) Work out the size of angle \(ADC\). (2 marks) (c) Work out the size of angle \(ABC\). Give a reason for your answer. (1 mark)

    A circle diagram showing a tangent and a cyclic quadrilateral.
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    Model answer

    (a) \(ACD = 52^\circ\), by the alternate segment theorem. (b) \(ADC = 180 - 52 - 43 = 85^\circ\). (c) \(ABC = 180 - 85 = 95^\circ\), because opposite angles of a cyclic quadrilateral add up to \(180^\circ\).

    Mark scheme

    • (a) \(52\) — B1
    • (a) Alternate segment theorem stated — C1
    • (b) \(180 - 52 - 43\) — M1
    • (b) \(85\) — A1
    • (c) \(95\) with the cyclic quadrilateral reason — A1
  6. 6 Prove [4 marks]

    \(A\), \(B\), \(C\) and \(D\) are points on a circle, centre \(O\), in that order around the circle. Prove that angle \(ABC\) and angle \(ADC\) add up to \(180^\circ\). (4 marks)

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    Model answer

    Let angle \(ABC = x\). The angle at the centre on the arc \(ADC\) is \(2x\), because the angle at the centre is twice the angle at the circumference. Let angle \(ADC = y\). The angle at the centre on the arc \(ABC\) is \(2y\). The angles round the point \(O\) add up to \(360^\circ\), so \(2x + 2y = 360\), which gives \(x + y = 180\).

    Mark scheme

    • Angle ABC = x, so the angle at the centre on arc ADC is 2x — B1
    • Angle ADC = y, so the angle at the centre on arc ABC is 2y — M1
    • Angles round a point: 2x + 2y = 360 — M1
    • Concludes x + y = 180 — C1

Quick check

  1. 1

    Why is a triangle made by two radii and a chord isosceles?

    1. AThe chord is a diameter
    2. BTwo of its sides are radii, so they are equal
    3. CAll three sides are radii
    4. DThe angles are all \(60^\circ\)
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    B: Two of its sides are radii, so they are equal

    Two radii are always equal.

  2. 2

    In a proof, what should be written next to every step?

    1. AA reason
    2. BA number
    3. CA diagram
    4. DA question
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    A: A reason

    Each step needs a reason.

  3. 3

    What does the exterior angle of a triangle equal?

    1. AThe third interior angle
    2. B\(90^\circ\)
    3. C\(360^\circ\) minus the interior angle
    4. DThe sum of the two opposite interior angles
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    D: The sum of the two opposite interior angles

    This is the exterior angle theorem.

  4. 4

    The radius to a point on a circle has gradient \(\dfrac{2}{3}\). What is the gradient of the tangent at that point?

    1. A\(\dfrac{3}{2}\)
    2. B\(-\dfrac{2}{3}\)
    3. C\(-\dfrac{3}{2}\)
    4. D\(\dfrac{2}{3}\)
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    C: \(-\dfrac{3}{2}\)

    A tangent is perpendicular to the radius, so its gradient is the negative reciprocal.

  5. 5

    What is the gradient of the radius from the origin to \((3, 4)\)?

    1. A\(\dfrac{3}{4}\)
    2. B\(\dfrac{4}{3}\)
    3. C\(-\dfrac{4}{3}\)
    4. D\(7\)
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    B: \(\dfrac{4}{3}\)

    \(\dfrac{4 - 0}{3 - 0} = \dfrac{4}{3}\).

  6. 6

    What is the gradient of the tangent to \(x^2 + y^2 = 25\) at \((3, 4)\)?

    1. A\(-\dfrac{3}{4}\)
    2. B\(\dfrac{4}{3}\)
    3. C\(\dfrac{3}{4}\)
    4. D\(-\dfrac{4}{3}\)
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    A: \(-\dfrac{3}{4}\)

    The radius has gradient \(\dfrac{4}{3}\), so the tangent has gradient \(-\dfrac{3}{4}\).

  7. 7

    \(AOC\) is \(150^\circ\) at the centre, and \(D\) is on the minor arc \(AC\). What is angle \(ADC\)?

    1. A\(75^\circ\)
    2. B\(150^\circ\)
    3. C\(210^\circ\)
    4. D\(105^\circ\)
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    D: \(105^\circ\)

    \(B\) on the major arc gives \(ABC = 75^\circ\), and \(ADC = 180 - 75 = 105^\circ\).

  8. 8

    Which line is the tangent to \(x^2 + y^2 = 25\) at \((3, 4)\)?

    1. A\(4x + 3y = 25\)
    2. B\(3x - 4y = 25\)
    3. C\(3x + 4y = 25\)
    4. D\(x + y = 7\)
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    C: \(3x + 4y = 25\)

    Gradient \(-\dfrac{3}{4}\) through \((3, 4)\) gives \(y - 4 = -\dfrac{3}{4}(x - 3)\), which rearranges to \(3x + 4y = 25\).

  9. 9

    Why must a proof not rely on measuring angles in a diagram?

    1. ADiagrams are always wrong
    2. BThe proof must work for every case, not just the one drawn
    3. CAngles cannot be measured
    4. DIt takes too long
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    B: The proof must work for every case, not just the one drawn

    A proof uses letters and reasons so that it works for any angle.