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WMA12/01 Oct 2024, Question 10: 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 given by x² + y² + 4x − 30y + 209 = 0, determine its centre and radius, then use the tangent line y = mx + 1 to derive an equation in m and find the possible coordinates of the tangency point P.

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

Part (ai) · 2 marks

Work out the centre coordinates of the circle from its expanded equation.

  1. 1Complete the square: (x + 2)² − 4 + (y − 15)² − 225 + 209 = 0
  2. 2(x + 2)² + (y − 15)² = 20
  3. 3Centre (−2, 15)

Answer(−2, 15)

Part (aii) · 1 mark

Calculate the radius of the circle, giving the exact value.

  1. 1From (x + 2)² + (y − 15)² = 20
  2. 2r² = 20, so r = √20 = 2√5

Answer√20 (= 2√5)

Part (b) · 3 marks

Using the fact that the line y = mx + 1 touches the circle at P, show that 2m² − 7m − 22 = 0.

  1. 1Substitute y = mx + 1: x² + (mx + 1)² + 4x − 30(mx + 1) + 209 = 0
  2. 2(1 + m²)x² + (4 − 28m)x + 180 = 0
  3. 3Tangent, so b² − 4ac = 0: (4 − 28m)² − 4(1 + m²)(180) = 0
  4. 416 − 224m + 784m² − 720 − 720m² = 0, so 64m² − 224m − 704 = 0
  5. 5Divide by 32: 2m² − 7m − 22 = 0

AnswerShown: (1 + m²)x² + (4 − 28m)x + 180 = 0; b² − 4ac = 0 gives 64m² − 224m − 704 = 0, i.e. 2m² − 7m − 22 = 0

Part (c) · 4 marks

Solve for m and then determine all possible coordinate pairs for the point P.

  1. 12m² − 7m − 22 = 0: (2m − 11)(m + 2) = 0, so m = 11/2 or m = −2
  2. 2m = 11/2: (125/4)x² − 150x + 180 = 0, a repeated root x = 12/5, so y = 11/2 × 12/5 + 1 = 71/5
  3. 3m = −2: 5x² + 60x + 180 = 0, a repeated root x = −6, so y = −2(−6) + 1 = 13
  4. 4P = (12/5, 71/5) or P = (−6, 13)

Answer(12/5, 71/5) and (−6, 13)

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Also searched as: WMA12 October 2024 · WMA12/01 · wma12 oct24 · IAS Maths Pure Mathematics 2 October 2024 · question 10.
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