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WMA12/01 Jun 2022, Question 10: Circles Meeting Lines, and Tangents

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 12 marks

The question, in short

For circle C centred at X(3, 5) with radius r and line l given by y = 2x + k, work through showing an intersection equation, a tangency condition, and finding k when AB = 2r.

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

Part (a) · 3 marks

Substitute the line equation into the circle equation and rearrange to show the given quadratic in x equals zero.

  1. 1Circle: (x − 3)² + (y − 5)² = r²
  2. 2Substitute y = 2x + k: (x − 3)² + (2x + k − 5)² = r²
  3. 3x² − 6x + 9 + 4x² + 4(k − 5)x + (k − 5)² = r²
  4. 45x² + (4k − 26)x + k² − 10k + 25 + 9 − r² = 0
  5. 55x² + (4k − 26)x + k² − 10k + 34 − r² = 0

AnswerShown: substituting y = 2x + k into (x − 3)² + (y − 5)² = r² and collecting terms gives 5x² + (4k − 26)x + k² − 10k + 34 − r² = 0

Part (b) · 3 marks

Using that l is tangent, set the discriminant of the quadratic to zero and rearrange to show 5r² = (k + p)², finding p.

  1. 1Tangent means one repeated root: b² − 4ac = 0
  2. 2(4k − 26)² − 4 × 5 × (k² − 10k + 34 − r²) = 0
  3. 316k² − 208k + 676 − 20k² + 200k − 680 + 20r² = 0
  4. 420r² = 4k² + 8k + 4, so 5r² = k² + 2k + 1
  5. 55r² = (k + 1)², so p = 1

AnswerShown: 5r² = (k + 1)², so p = 1

Part (c) · 6 marks

With A on the y-axis, B the tangency point and AB = 2r, use the geometry in Figure 2 to find the value of k.

  1. 1A is (0, k). Triangle AXB has a right angle at B (tangent ⟂ radius), so AB² + r² = XA²
  2. 2XA² = (3 − 0)² + (5 − k)² = 9 + (5 − k)²
  3. 3AB = 2r, so AB² + r² = 4r² + r² = 5r²
  4. 45r² = 9 + (5 − k)²
  5. 5Using (b): (k + 1)² = 9 + (5 − k)², so k² + 2k + 1 = 34 − 10k + k²
  6. 612k = 33, so k = 11/4

Answerk = 11/4

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Also searched as: WMA12 June 2022 · WMA12/01 · wma12 jun22 · IAS Maths Pure Mathematics 2 June 2022 · question 10.
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