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WMA12/01 Jan 2022, Question 6: Chords and Tangent Geometry

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 8 marks

The question, in short

For circle C through P(23, 14), Q(15, −30) and R(−7, −26), prove angle PQR is 90°, then find the circle's centre and radius, and the tangent at the point S giving the greatest QS distance.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 2 marks

Show that angle PQR equals 90° using the given coordinates of P, Q and R.

  1. 1Gradient PQ = (14 + 30)/(23 − 15) = 44/8 = 11/2
  2. 2Gradient QR = (−30 + 26)/(15 + 7) = −4/22 = −2/11
  3. 3(11/2) × (−2/11) = −1, so PQ is perpendicular to QR and angle PQR = 90°

AnswerShown: gradient PQ × gradient QR = (11/2) × (−2/11) = −1, so angle PQR = 90°

Part (bi) · 1 mark

Deduce the centre of circle C.

  1. 1Angle PQR = 90°, so PR is a diameter
  2. 2Centre = midpoint of PR = ((23 − 7)/2, (14 − 26)/2) = (8, −6)

Answer(8, −6)

Part (bii) · 2 marks

Find the radius of circle C.

  1. 1r² = (23 − 8)² + (14 + 6)² = 225 + 400 = 625
  2. 2r = 25

Answerr = 25

Part (c) · 3 marks

With S on C giving the greatest distance QS, find the tangent to C at S in the form ax + by + c = 0 with integer coefficients.

  1. 1QS is greatest when QS is a diameter, so S = 2(8, −6) − (15, −30) = (1, 18)
  2. 2Gradient of radius CS = (18 + 6)/(1 − 8) = −24/7, so tangent gradient = 7/24
  3. 3y − 18 = (7/24)(x − 1)
  4. 424y − 432 = 7x − 7, so 7x − 24y + 425 = 0

Answer7x − 24y + 425 = 0

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Also searched as: WMA12 January 2022 · WMA12/01 · wma12 jan22 · IAS Maths Pure Mathematics 2 January 2022 · question 6.
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