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Exam questions · Maths

Circle Theorems

  • 30 exam questions
  • 102 marks
  • 45 quick checks

Angles at the Centre and in a Semicircle

Just this lesson
  1. 1 Work out [2 marks]

    Diagram NOT accurately drawn. \(A\), \(B\) and \(C\) are points on a circle, centre \(O\). Angle \(AOB = 124^\circ\). Work out the size of angle \(ACB\). Give a reason for your answer. (2 marks)

    A circle diagram showing the angle at the centre and the angle at the circumference.
    Show answerHide answer

    Model answer

    Angle \(ACB = 62^\circ\), because the angle at the centre is twice the angle at the circumference.

    Mark scheme

    • \(62\) — B1
    • The angle at the centre is twice the angle at the circumference — C1
  2. 2 Work out [3 marks]

    Diagram NOT accurately drawn. \(AB\) is a diameter of a circle and \(C\) is a point on the circle. Angle \(BAC = 34^\circ\). Work out the size of angle \(ABC\). Give a reason for each step of your working. (3 marks)

    A circle diagram showing a triangle in a semicircle.
    Show answerHide answer

    Model answer

    Angle \(ACB = 90^\circ\), because the angle in a semicircle is a right angle. Then \(ABC = 180 - 90 - 34 = 56^\circ\).

    Mark scheme

    • Angle \(ACB = 90^\circ\) — B1
    • The angle in a semicircle is a right angle — C1
    • \(56\) — B1
  3. 3 Work out [3 marks]

    \(A\), \(B\) and \(C\) are points on a circle, centre \(O\), with \(C\) on the major arc \(AB\). Angle \(AOB = 6x + 20\) and angle \(ACB = 2x + 14\). Work out the value of \(x\). (3 marks)

    Show answerHide answer

    Model answer

    The angle at the centre is twice the angle at the circumference, so \(6x + 20 = 2(2x + 14) = 4x + 28\). Then \(2x = 8\) and \(x = 4\).

    Mark scheme

    • \(6x + 20 = 2(2x + 14)\) — M1
    • Expands and collects terms, for example \(2x = 8\) — M1
    • \(4\) — A1
  4. 4 Work out [4 marks]

    \(A\), \(B\) and \(C\) are points on a circle, centre \(O\), with \(C\) on the major arc \(AB\). Angle \(OAB = 35^\circ\). Work out the size of angle \(ACB\). Give reasons for your answer. (4 marks)

    Show answerHide answer

    Model answer

    \(OA = OB\) because they are radii, so triangle \(OAB\) is isosceles and angle \(OBA = 35^\circ\). Then \(AOB = 180 - 35 - 35 = 110^\circ\). The angle at the centre is twice the angle at the circumference, so \(ACB = 110 \div 2 = 55^\circ\).

    Mark scheme

    • \(180 - 35 - 35\) or angle \(OBA = 35^\circ\) — M1
    • Angle \(AOB = 110^\circ\) — A1
    • \(55\) — A1
    • Reasons: radii make an isosceles triangle, and the angle at the centre is twice the angle at the circumference — C1
  5. 5 Work out [4 marks]

    \(AB\) is a diameter of a circle and \(C\) is a point on the circle. \(AC = 8\) cm and \(BC = 15\) cm. Work out the radius of the circle. Give a reason for your answer. (4 marks)

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    Model answer

    Angle \(ACB = 90^\circ\), because the angle in a semicircle is a right angle. \(AB^2 = 8^2 + 15^2 = 64 + 225 = 289\), so \(AB = 17\) cm and the radius is \(8.5\) cm.

    Mark scheme

    • Angle in a semicircle is a right angle — C1
    • \(8^2 + 15^2\) or \(64 + 225\) — M1
    • \(AB = 17\) — A1
    • \(8.5\) — B1
  6. 6 Work out [3 marks]

    Diagram NOT accurately drawn. \(A\), \(B\) and \(C\) are points on a circle, centre \(O\). Angle \(AOB = 124^\circ\), and \(C\) is on the minor arc \(AB\). Work out the size of angle \(ACB\). Give a reason for your answer. (3 marks)

    A circle diagram showing the angle at the circumference on the minor arc.
    Show answerHide answer

    Model answer

    The reflex angle \(AOB\) is \(360 - 124 = 236^\circ\). The angle at the circumference is half the angle at the centre, so \(ACB = 236 \div 2 = 118^\circ\).

    Mark scheme

    • \(360 - 124 = 236\) — M1
    • \(118\) — A1
    • The angle at the centre is twice the angle at the circumference — C1

Quick check

  1. 1

    The angle at the centre of a circle is \(84^\circ\). What is the angle at the circumference made by the same arc?

    1. A\(168^\circ\)
    2. B\(42^\circ\)
    3. C\(84^\circ\)
    4. D\(96^\circ\)
    Show answerHide answer

    B: \(42^\circ\)

    The angle at the centre is twice the angle at the circumference, so the angle at the circumference is \(84 \div 2 = 42^\circ\).

  2. 2

    What is the size of an angle in a semicircle?

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

    A: \(90^\circ\)

    The angle at the centre on a diameter is \(180^\circ\), so the angle at the circumference is half of it, \(90^\circ\).

  3. 3

    Which reason fits this statement? \(\angle AOB = 2 \times \angle ACB\)

    1. AAngles in the same segment are equal
    2. BThe angle in a semicircle is \(90^\circ\)
    3. COpposite angles of a cyclic quadrilateral add up to \(180^\circ\)
    4. DThe angle at the centre is twice the angle at the circumference
    Show answerHide answer

    D: The angle at the centre is twice the angle at the circumference

    The angle at \(O\), the centre, is double the angle at \(C\), on the circumference.

  4. 4

    \(AB\) is a diameter and \(C\) is on the circle. Angle \(CAB = 35^\circ\). What is angle \(CBA\)?

    1. A\(35^\circ\)
    2. B\(90^\circ\)
    3. C\(55^\circ\)
    4. D\(145^\circ\)
    Show answerHide answer

    C: \(55^\circ\)

    Angle \(ACB = 90^\circ\) in a semicircle, so \(CBA = 180 - 90 - 35 = 55^\circ\).

  5. 5

    The angle at the circumference is \(3x\) and the angle at the centre on the same arc is \(5x + 20\). What is \(x\)?

    1. A\(10\)
    2. B\(20\)
    3. C\(4\)
    4. D\(40\)
    Show answerHide answer

    B: \(20\)

    \(5x + 20 = 2 \times 3x = 6x\), so \(x = 20\).

  6. 6

    In triangle \(OAB\), \(O\) is the centre and \(\angle OAB = 28^\circ\). What is angle \(AOB\)?

    1. A\(124^\circ\)
    2. B\(62^\circ\)
    3. C\(56^\circ\)
    4. D\(152^\circ\)
    Show answerHide answer

    A: \(124^\circ\)

    \(OA = OB\) are radii, so \(\angle OBA = 28^\circ\) and \(\angle AOB = 180 - 28 - 28 = 124^\circ\).

  7. 7

    \(AB\) is a diameter of length 10 cm and \(AC = 6\) cm. What is \(BC\)?

    1. A4 cm
    2. B\(\sqrt{136}\) cm
    3. C16 cm
    4. D8 cm
    Show answerHide answer

    D: 8 cm

    Angle \(ACB = 90^\circ\) in a semicircle, so \(BC^2 = 10^2 - 6^2 = 64\) and \(BC = 8\).

  8. 8

    Two points \(A\) and \(B\) are on a circle with centre \(O\), and \(\angle AOB = 130^\circ\). \(C\) is on the minor arc \(AB\). What is angle \(ACB\)?

    1. A\(65^\circ\)
    2. B\(130^\circ\)
    3. C\(115^\circ\)
    4. D\(230^\circ\)
    Show answerHide answer

    C: \(115^\circ\)

    The reflex angle at the centre is \(360 - 130 = 230^\circ\), so \(ACB = 230 \div 2 = 115^\circ\).

  9. 9

    Which statement is true for every triangle with a diameter as one side and its third corner on the circle?

    1. AIt is isosceles
    2. BIt is right-angled
    3. CIt is equilateral
    4. DIt has an angle of \(60^\circ\)
    Show answerHide answer

    B: It is right-angled

    The angle opposite the diameter is in a semicircle, so it is always \(90^\circ\).

Same Segment and Cyclic Quadrilaterals

Just this lesson
  1. 1 Work out [2 marks]

    Diagram NOT accurately drawn. \(A\), \(B\), \(C\) and \(D\) are points on a circle. Angle \(ACB = 35^\circ\). Work out the size of angle \(ADB\). Give a reason for your answer. (2 marks)

    A circle diagram showing two angles in the same segment.
    Show answerHide answer

    Model answer

    Angle \(ADB = 35^\circ\), because angles in the same segment are equal.

    Mark scheme

    • \(35\) — B1
    • Angles in the same segment are equal — C1
  2. 2 Work out [4 marks]

    Diagram NOT accurately drawn. \(ABCD\) is a cyclic quadrilateral. Angle \(ABC = 77^\circ\) and angle \(BAD = 100^\circ\). (a) Work out the size of angle \(ADC\). Give a reason for your answer. (2 marks) (b) Work out the size of angle \(BCD\). Give a reason for your answer. (2 marks)

    A circle diagram showing a cyclic quadrilateral with two angles given.
    Show answerHide answer

    Model answer

    (a) \(ADC = 180 - 77 = 103^\circ\), because opposite angles of a cyclic quadrilateral add up to \(180^\circ\). (b) \(BCD = 180 - 100 = 80^\circ\), for the same reason.

    Mark scheme

    • (a) \(103\) — B1
    • (a) Opposite angles of a cyclic quadrilateral add up to 180 degrees — C1
    • (b) \(80\) — B1
    • (b) The same reason given — C1
  3. 3 Work out [3 marks]

    \(ABCD\) is a cyclic quadrilateral. Angle \(A = 3x + 5\) and angle \(C = 2x + 25\). Work out the value of \(x\). (3 marks)

    Show answerHide answer

    Model answer

    Opposite angles add up to \(180^\circ\), so \(3x + 5 + 2x + 25 = 180\). Then \(5x = 150\) and \(x = 30\).

    Mark scheme

    • \((3x + 5) + (2x + 25) = 180\) — M1
    • \(5x + 30 = 180\) or \(5x = 150\) — M1
    • \(30\) — A1
  4. 4 Work out [3 marks]

    Diagram NOT accurately drawn. \(ABCD\) is a cyclic quadrilateral. The side \(AB\) is extended to the point \(E\). Angle \(CBE = 68^\circ\). (a) Work out the size of angle \(ABC\). (1 mark) (b) Work out the size of angle \(ADC\). Give a reason for your answer. (2 marks)

    A circle diagram showing a cyclic quadrilateral with a side extended.
    Show answerHide answer

    Model answer

    (a) Angles on a straight line add up to \(180^\circ\), so \(ABC = 180 - 68 = 112^\circ\). (b) \(ADC = 180 - 112 = 68^\circ\), because opposite angles of a cyclic quadrilateral add up to \(180^\circ\).

    Mark scheme

    • (a) \(112\) — B1
    • (b) \(68\) — B1
    • (b) Opposite angles of a cyclic quadrilateral add up to 180 degrees — C1
  5. 5 Work out [4 marks]

    \(A\), \(B\), \(C\) and \(D\) are points on a circle. The lines \(AC\) and \(BD\) cross at \(E\). Angle \(CAD = 35^\circ\) and angle \(ABC = 100^\circ\). (a) Work out the size of angle \(CBD\). Give a reason for your answer. (2 marks) (b) Work out the size of angle \(ADC\). Give a reason for your answer. (2 marks)

    Show answerHide answer

    Model answer

    (a) \(CBD = 35^\circ\), because angles in the same segment are equal. (b) \(ADC = 180 - 100 = 80^\circ\), because opposite angles of a cyclic quadrilateral add up to \(180^\circ\).

    Mark scheme

    • (a) \(35\) — B1
    • (a) Angles in the same segment are equal — C1
    • (b) \(80\) — B1
    • (b) Opposite angles of a cyclic quadrilateral add up to 180 degrees — C1
  6. 6 Show that [2 marks]

    \(PQRS\) is a quadrilateral. Angle \(P = 105^\circ\) and angle \(R = 80^\circ\). Show that \(PQRS\) cannot be a cyclic quadrilateral. (2 marks)

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    Model answer

    \(105 + 80 = 185\), which is not \(180^\circ\). The opposite angles of a cyclic quadrilateral add up to \(180^\circ\), so \(PQRS\) cannot be cyclic.

    Mark scheme

    • \(105 + 80 = 185\) — M1
    • States that this is not 180 degrees, so the quadrilateral cannot be cyclic — C1

Quick check

  1. 1

    Two angles are in the same segment of a circle. One is \(47^\circ\). What is the other?

    1. A\(133^\circ\)
    2. B\(94^\circ\)
    3. C\(47^\circ\)
    4. D\(43^\circ\)
    Show answerHide answer

    C: \(47^\circ\)

    Angles in the same segment are equal.

  2. 2

    What do opposite angles of a cyclic quadrilateral add up to?

    1. A\(360^\circ\)
    2. B\(180^\circ\)
    3. C\(90^\circ\)
    4. D\(270^\circ\)
    Show answerHide answer

    B: \(180^\circ\)

    This is the cyclic quadrilateral theorem.

  3. 3

    A cyclic quadrilateral has an angle of \(112^\circ\). What is the opposite angle?

    1. A\(68^\circ\)
    2. B\(112^\circ\)
    3. C\(248^\circ\)
    4. D\(78^\circ\)
    Show answerHide answer

    A: \(68^\circ\)

    \(180 - 112 = 68\).

  4. 4

    What is a cyclic quadrilateral?

    1. AA quadrilateral with all sides equal
    2. BA quadrilateral with a circle inside it
    3. CA quadrilateral with four right angles
    4. DA quadrilateral with all four corners on a circle
    Show answerHide answer

    D: A quadrilateral with all four corners on a circle

    “Cyclic” means all the corners lie on one circle.

  5. 5

    In a cyclic quadrilateral \(ABCD\), \(\angle A = 2x + 10\) and \(\angle C = 3x + 20\). What is \(x\)?

    1. A\(10\)
    2. B\(15\)
    3. C\(30\)
    4. D\(50\)
    Show answerHide answer

    C: \(30\)

    \(5x + 30 = 180\), so \(5x = 150\) and \(x = 30\).

  6. 6

    \(ABCD\) is cyclic and the side \(AB\) is extended to \(E\). Angle \(CBE = 70^\circ\). What is angle \(ADC\)?

    1. A\(110^\circ\)
    2. B\(70^\circ\)
    3. C\(35^\circ\)
    4. D\(140^\circ\)
    Show answerHide answer

    B: \(70^\circ\)

    An exterior angle of a cyclic quadrilateral equals the interior opposite angle.

  7. 7

    Which of these must be true for the angles \(ACB\) and \(ADB\) to be equal?

    1. A\(C\) and \(D\) are on the same side of the chord \(AB\)
    2. B\(C\) and \(D\) are on opposite sides of \(AB\)
    3. C\(AB\) is a diameter
    4. D\(C\), \(D\) and \(O\) are in a line
    Show answerHide answer

    A: \(C\) and \(D\) are on the same side of the chord \(AB\)

    Angles in the same segment are made on the same side of a chord.

  8. 8

    A quadrilateral has opposite angles of \(95^\circ\) and \(80^\circ\). Can it be cyclic?

    1. AYes, because they are both less than \(180^\circ\)
    2. BYes, because all quadrilaterals are cyclic
    3. CNo, because the angles must be equal
    4. DNo, because \(95 + 80 \ne 180\)
    Show answerHide answer

    D: No, because \(95 + 80 \ne 180\)

    Opposite angles of a cyclic quadrilateral must add up to \(180^\circ\), and \(95 + 80 = 175\).

  9. 9

    \(A\), \(B\), \(C\) and \(D\) are on a circle, with \(AC\) and \(BD\) meeting at \(E\). Angle \(CAD = 36^\circ\). What is angle \(CBD\)?

    1. A\(72^\circ\)
    2. B\(144^\circ\)
    3. C\(36^\circ\)
    4. D\(54^\circ\)
    Show answerHide answer

    C: \(36^\circ\)

    Angles \(CAD\) and \(CBD\) are made by the chord \(CD\) on the same side, so they are equal.

Tangents and Chords

Just this lesson
  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.
    Show answerHide answer

    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)

    Show answerHide answer

    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.
    Show answerHide answer

    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.
    Show answerHide answer

    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)

    Show answerHide answer

    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)

    Show answerHide answer

    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\(90^\circ\)
    Show answerHide answer

    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
    Show answerHide answer

    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
    Show answerHide answer

    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
    Show answerHide answer

    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\(80^\circ\)
    Show answerHide answer

    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
    Show answerHide answer

    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\(40^\circ\)
    Show answerHide answer

    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
    Show answerHide answer

    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\(\sqrt{d^2 - r^2}\)
    Show answerHide answer

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

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

The Alternate Segment Theorem

Just this lesson
  1. 1 Work out [2 marks]

    Diagram NOT accurately drawn. \(TAS\) is a tangent to the circle at \(A\). \(B\) and \(C\) are points on the circle. Angle \(SAB = 48^\circ\). Work out the size of angle \(ACB\). Give a reason for your answer. (2 marks)

    A circle diagram showing a tangent, a chord and an angle in the alternate segment.
    Show answerHide answer

    Model answer

    Angle \(ACB = 48^\circ\), because the angle between a tangent and a chord equals the angle in the alternate segment.

    Mark scheme

    • \(48\) — B1
    • The angle between a tangent and a chord equals the angle in the alternate segment — C1
  2. 2 Work out [4 marks]

    Diagram NOT accurately drawn. \(TAS\) is a tangent to the circle at \(A\). \(B\) and \(C\) are points on the circle. Angle \(TAB = 52^\circ\) and angle \(ABC = 65^\circ\). Work out the size of angle \(BAC\). Give reasons for your answer. (4 marks)

    A circle diagram showing a tangent and a triangle ABC.
    Show answerHide answer

    Model answer

    Angle \(ACB = 52^\circ\), because the angle between a tangent and a chord equals the angle in the alternate segment. Then \(BAC = 180 - 52 - 65 = 63^\circ\), because the angles in a triangle add up to \(180^\circ\).

    Mark scheme

    • \(ACB = 52\) — B1
    • Alternate segment theorem stated — C1
    • \(180 - 52 - 65\) — M1
    • \(63\) — A1
  3. 3 Work out [3 marks]

    \(TA\) is a tangent to a circle at \(A\), and \(B\) and \(C\) are points on the circle, with \(C\) in the alternate segment. Angle \(TAB = 3x + 10\) and angle \(ACB = 5x - 14\). Work out the value of \(x\). (3 marks)

    Show answerHide answer

    Model answer

    By the alternate segment theorem, \(3x + 10 = 5x - 14\). Then \(24 = 2x\), so \(x = 12\).

    Mark scheme

    • \(3x + 10 = 5x - 14\) — M1
    • \(2x = 24\) — M1
    • \(12\) — A1
  4. 4 Work out [4 marks]

    Diagram NOT accurately drawn. \(PA\) and \(PB\) are tangents to a circle, centre \(O\). \(C\) is a point on the circle. Angle \(APB = 64^\circ\). Work out the size of angle \(ACB\). Give reasons for your answer. (4 marks)

    A circle diagram showing two tangents from P and a point C on the circle.
    Show answerHide answer

    Model answer

    \(PA = PB\), so triangle \(PAB\) is isosceles and \(PAB = (180 - 64) \div 2 = 58^\circ\). By the alternate segment theorem, \(ACB = PAB = 58^\circ\).

    Mark scheme

    • \((180 - 64) \div 2\) — M1
    • \(PAB = 58\) — A1
    • \(ACB = 58\) — B1
    • Alternate segment theorem stated — C1
  5. 5 Work out [4 marks]

    \(AD\) is a diameter of a circle. \(TA\) is a tangent to the circle at \(A\). \(B\) is a point on the circle. Angle \(TAB = 38^\circ\). (a) Work out the size of angle \(ADB\). Give a reason for your answer. (2 marks) (b) Work out the size of angle \(DAB\). (2 marks)

    Show answerHide answer

    Model answer

    (a) \(ADB = 38^\circ\), by the alternate segment theorem. (b) \(TAD = 90^\circ\) because a tangent is perpendicular to the radius, so \(DAB = 90 - 38 = 52^\circ\).

    Mark scheme

    • (a) \(38\) — B1
    • (a) Alternate segment theorem stated — C1
    • (b) \(90 - 38\), using angle \(TAD = 90^\circ\) — M1
    • (b) \(52\) — A1
  6. 6 Prove [4 marks]

    \(TA\) is a tangent to a circle at \(A\). \(AD\) is a diameter. \(B\) is a point on the circle. Prove that angle \(TAB\) equals angle \(ADB\). (4 marks)

    Show answerHide answer

    Model answer

    Angle \(TAD = 90^\circ\), because a tangent is perpendicular to the radius. Angle \(ABD = 90^\circ\), because the angle in a semicircle is \(90^\circ\). So \(TAB = 90 - BAD\), and in triangle \(ABD\), \(ADB = 90 - BAD\). Therefore \(TAB = ADB\).

    Mark scheme

    • Angle \(TAD = 90^\circ\), because a tangent is perpendicular to the radius — B1
    • Angle \(ABD = 90^\circ\), because the angle in a semicircle is a right angle — B1
    • \(TAB = 90 - BAD\) and \(ADB = 90 - BAD\) — M1
    • Concludes that the angles are equal — C1

Quick check

  1. 1

    What does the alternate segment theorem say?

    1. AThe angle between a tangent and a chord equals the angle in the alternate segment
    2. BThe angle between a tangent and a chord is \(90^\circ\)
    3. CAngles in the same segment are equal
    4. DThe angle in a semicircle is \(90^\circ\)
    Show answerHide answer

    A: The angle between a tangent and a chord equals the angle in the alternate segment

    This is the alternate segment theorem.

  2. 2

    Where must the chord start for the alternate segment theorem to apply?

    1. AAt the centre of the circle
    2. BAnywhere on the circle
    3. CAt the end of a diameter
    4. DAt the point of contact of the tangent
    Show answerHide answer

    D: At the point of contact of the tangent

    The chord starts at the point where the tangent touches the circle.

  3. 3

    The angle between a tangent and a chord is \(58^\circ\). What is the angle in the alternate segment?

    1. A\(32^\circ\)
    2. B\(116^\circ\)
    3. C\(58^\circ\)
    4. D\(122^\circ\)
    Show answerHide answer

    C: \(58^\circ\)

    They are equal.

  4. 4

    What does “alternate” mean in the alternate segment theorem?

    1. AOpposite the centre
    2. BOn the other side of the chord
    3. CEvery other segment
    4. DInside the triangle
    Show answerHide answer

    B: On the other side of the chord

    The alternate segment is the one on the opposite side of the chord from the angle.

  5. 5

    \(TA\) is a tangent at \(A\). Angle \(TAB = 40^\circ\) and angle \(ABC = 75^\circ\), where \(C\) is in the alternate segment. What is angle \(BAC\)?

    1. A\(65^\circ\)
    2. B\(40^\circ\)
    3. C\(75^\circ\)
    4. D\(115^\circ\)
    Show answerHide answer

    A: \(65^\circ\)

    \(ACB = 40^\circ\) by the alternate segment theorem, so \(BAC = 180 - 75 - 40 = 65^\circ\).

  6. 6

    \(PA\) and \(PB\) are tangents and \(\angle APB = 64^\circ\). \(C\) is on the major arc. What is angle \(ACB\)?

    1. A\(64^\circ\)
    2. B\(116^\circ\)
    3. C\(32^\circ\)
    4. D\(58^\circ\)
    Show answerHide answer

    D: \(58^\circ\)

    \(PAB = (180 - 64) \div 2 = 58^\circ\), and this equals the angle in the alternate segment.

  7. 7

    The angle between the tangent and the chord is \(2x + 4\) and the angle in the alternate segment is \(3x - 10\). What is \(x\)?

    1. A\(6\)
    2. B\(32\)
    3. C\(14\)
    4. D\(7\)
    Show answerHide answer

    C: \(14\)

    \(2x + 4 = 3x - 10\), so \(x = 14\).

  8. 8

    \(AD\) is a diameter, \(TA\) is a tangent at \(A\) and \(B\) is on the circle. Angle \(TAB = 36^\circ\). What is angle \(DAB\)?

    1. A\(36^\circ\)
    2. B\(54^\circ\)
    3. C\(90^\circ\)
    4. D\(144^\circ\)
    Show answerHide answer

    B: \(54^\circ\)

    \(TAD = 90^\circ\) because a tangent is perpendicular to the radius, so \(DAB = 90 - 36 = 54^\circ\).

  9. 9

    A tangent at \(A\) makes an angle of \(x\) with the chord \(AB\). What is the angle \(AOB\) at the centre, on the same side as that angle?

    1. A\(2x\)
    2. B\(x\)
    3. C\(90 - x\)
    4. D\(180 - x\)
    Show answerHide answer

    A: \(2x\)

    The angle in the alternate segment is \(x\), and the angle at the centre is twice that.

Circle Theorem Proofs and Problems

Just this lesson
  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.
    Show answerHide answer

    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.
    Show answerHide answer

    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)

    Show answerHide answer

    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)

    Show answerHide answer

    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.
    Show answerHide answer

    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)

    Show answerHide answer

    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 \(60^\circ\)
    Show answerHide answer

    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
    Show answerHide answer

    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
    Show answerHide answer

    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\(\dfrac{2}{3}\)
    Show answerHide answer

    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\(7\)
    Show answerHide answer

    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\(-\dfrac{4}{3}\)
    Show answerHide answer

    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\(105^\circ\)
    Show answerHide answer

    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\(x + y = 7\)
    Show answerHide answer

    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
    Show answerHide answer

    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.