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WMA12/01 Jan 2021, Question 9: Circles Meeting Lines, and Tangents

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 10 marks

The question, in short

For a circle with equation (x − k)² + (y − 2k)² = k + 7, find centre and radius in terms of k, use point P(2, 3) to form and solve a quadratic in k, then find the exact area of triangle OPT for k < 2.

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

Part (ai) · 1 mark

State the centre coordinates of circle C in terms of k.

  1. 1Compare with (x − a)² + (y − b)² = r²
  2. 2Centre = (k, 2k)

Answer(k, 2k)

Part (aii) · 1 mark

State the radius of circle C in terms of k.

  1. 1r² = k + 7
  2. 2Radius = √(k + 7)

Answer√(k + 7)

Part (bi) · 2 marks

Substitute P(2, 3) into the circle equation and show it simplifies to 5k² − 17k + 6 = 0.

  1. 1Substitute (2, 3): (2 − k)² + (3 − 2k)² = k + 7
  2. 24 − 4k + k² + 9 − 12k + 4k² = k + 7
  3. 35k² − 16k + 13 = k + 7
  4. 45k² − 17k + 6 = 0

AnswerShown: (2 − k)² + (3 − 2k)² = k + 7 simplifies to 5k² − 17k + 6 = 0

Part (bii) · 1 mark

Solve 5k² − 17k + 6 = 0 to find the possible values of k.

  1. 1(5k − 2)(k − 3) = 0
  2. 2k = 2/5 or k = 3

Answerk = 2/5 or k = 3

Part (c) · 5 marks

Using k < 2, find where the tangent at P meets the x-axis at T, then calculate the exact area of triangle OPT.

  1. 1k < 2, so k = 2/5 and the centre is (2/5, 4/5)
  2. 2Gradient of radius to P(2, 3) = (3 − 4/5)/(2 − 2/5) = (11/5)/(8/5) = 11/8
  3. 3Gradient of tangent = −8/11
  4. 4Tangent: y − 3 = −8/11 (x − 2); at y = 0, x = 2 + 33/8 = 49/8, so T = (49/8, 0)
  5. 5Area OPT = ½ × OT × height = ½ × 49/8 × 3 = 147/16

Answer147/16

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Also searched as: WMA12 January 2021 · WMA12/01 · wma12 jan21 · IAS Maths Pure Mathematics 2 January 2021 · question 9.
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