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IGCSE Maths · Edexcel 4XMA1 · Unit 2 Higher · Circle properties

Intersecting Chords and Secants — Worked Solutions

Full worked solutions to all 12 questions (35 marks), set out the way an Edexcel examiner wants to see them.

Section A — Warm-up (grades 6–7)
  1. (2 marks)Chords AB and CD meet at P. AP = 4 cm, PB = 6 cm and CP = 3 cm. Find PD.
    ABCDP463PD
    Solution
    1. 4 × 6 = 3 × PD
    2. PD = 24 ÷ 3
    8 cm
  2. (2 marks)Chords AB and CD meet at P. AP = 5 cm, PB = x cm, CP = 4 cm and PD = 10 cm. Find x.
    Solution
    1. 5 × x = 4 × 10
    2. x = 40 ÷ 5
    x = 8
  3. (3 marks)Two lines from P meet the circle at A and B, and at C and D. PA = 4 cm, AB = 6 cm and PC = 5 cm. Find CD.
    ABCDP465CD
    Solution
    1. PB = 4 + 6 = 10.
    2. 4 × 10 = 5 × PD → PD = 8.
    3. CD = 8 − 5 = 3.
    3 cm
  4. (2 marks)PT is a tangent to the circle. A line from P meets the circle at A and B. PA = 3 cm and AB = 9 cm. Find PT.
    ABTP39PT
    Solution
    1. PB = 12.
    2. PT² = 3 × 12 = 36, so PT = 6.
    6 cm
Section B — Core (grades 7–8)
  1. (3 marks)Chords AB and CD meet at P. AP = x cm, PB = (x + 5) cm, CP = 4 cm and PD = 6 cm. Find x.
    Solution
    1. x(x + 5) = 24 → x² + 5x − 24 = 0.
    2. (x + 8)(x − 3) = 0.
    3. x must be positive.
    x = 3
  2. (2 marks)PT is a tangent of length 8 cm. A line from P meets the circle at A and then B, with PA = 4 cm. Find AB.
    Solution
    1. PT² = PA × PB → 64 = 4 × PB → PB = 16.
    2. AB = 16 − 4 = 12.
    12 cm
  3. (3 marks)Two lines from P meet the circle at A and B, and at C and D. PA = x cm, AB = 7 cm, PC = 6 cm and CD = 4 cm. Find x.
    Solution
    1. PB = x + 7 and PD = 6 + 4 = 10.
    2. x(x + 7) = 60 → x² + 7x − 60 = 0.
    3. (x + 12)(x − 5) = 0, so x = 5.
    x = 5
  4. (4 marks)A diameter CD of a circle is perpendicular to a chord AB and meets it at P. AB = 16 cm and PC = 4 cm, where C is the end of the diameter nearer to AB. Find the radius of the circle.
    Solution
    1. The perpendicular from the centre bisects a chord, so AP = PB = 8.
    2. 8 × 8 = 4 × PD → PD = 16.
    3. CD = 4 + 16 = 20, so radius = 10.
    10 cm
Section C — Stretch (grades 8–9)
  1. (3 marks)A, B, C and D lie on a circle with centre O. ABCD is a cyclic quadrilateral with angle ABC = 104°, and B lies on the minor arc AC. Find (a) angle ADC (b) angle AOC. Give reasons.
    Solution
    1. (a) ADC = 180 − 104 = 76°, because opposite angles of a cyclic quadrilateral add up to 180°.
    2. (b) AOC = 2 × 76 = 152°, because the angle at the centre is twice the angle at the circumference.
    (a) 76° (b) 152°
  2. (3 marks)TA is a tangent to the circle at A. B and C are points on the circle. The angle between TA and the chord AB is 58°, and angle BAC = 70°. C is in the alternate segment. Find angle ABC. Give reasons.
    Solution
    1. Angle ACB = 58°: the angle between a tangent and a chord equals the angle in the alternate segment.
    2. Angle ABC = 180 − 70 − 58 = 52°: angles in a triangle add up to 180°.
    52°
  3. (4 marks)Chords AB and CD meet at P inside a circle. By considering triangles APC and DPB, prove that AP × PB = CP × PD.
    Solution
    1. Angle APC = angle DPB (vertically opposite angles).
    2. Angle CAB = angle CDB (angles in the same segment, both standing on arc CB).
    3. So triangles APC and DPB are similar (two equal angles).
    4. Corresponding sides are in proportion: AP/DP = CP/BP, so AP × PB = CP × PD.
    Proved by similar triangles
  4. (4 marks)P is 13 cm from the centre O of a circle of radius 5 cm. PT is a tangent. The line PO meets the circle at A and B, with A nearer to P.
    (a) Find PT. (b) Verify that PT² = PA × PB.
    Solution
    1. (a) Angle OTP = 90° (tangent ⟂ radius), so PT² = 13² − 5² = 144 and PT = 12.
    2. (b) PA = 13 − 5 = 8 and PB = 13 + 5 = 18.
    3. PA × PB = 144 = PT² ✓
    (a) 12 cm (b) 8 × 18 = 144 = 12²
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