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ExamsJEE AdvancedMaths

Find the equation of the circle of minimum radius that contains the three circles S1: x² + y² - 4y - 5 = 0, S2: x² + y² + 12x + 4y + 31 = 0, and S3: x² + y² + 6x + 12y + 36 = 0.

  1. (x + 31/18)² + (y + 23/12)² = 53.0 (approx)
  2. (x + 3)² + (y + 2)² = 49
  3. (x + 31/18)² + (y + 23/12)² = 18.3 (approx)
  4. (x - 31/18)² + (y - 23/12)² = 53.0 (approx)

Correct answer: (x + 31/18)² + (y + 23/12)² = 53.0 (approx)

Solution

The three circles have centres (0,2), (-6,-2), (-3,-6), each of radius 3; their circumcentre is (-31/18, -23/12), at distance ~4.28 from each centre, so the minimum enclosing radius is ~7.28 (R² ~ 53).

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