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ExamsJEE MainMaths

Find the equation of the circle passing through the intersection points of x² + y² - 1 = 0 and x² + y² - 2x - 4y + 1 = 0, and touching the line x + 2y = 0.

  1. x² + y² + x + 2y = 0
  2. x² + y² - x + 2y = 0
  3. x² + y² - x - 2y = 0
  4. 2(x² + y²) - 2x - 2y = 0

Correct answer: x² + y² - x - 2y = 0

Solution

Forming the family circle and applying the tangency condition (distance from centre to line = radius) gives the circle x² + y² - x - 2y = 0, which touches x + 2y = 0 at the origin.

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