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ExamsJEE MainMaths

On the segment joining A(0, 0) and B(3a, 0), choose points P and Q so that AP = PQ = QB. Three circles are then constructed with AP, PQ, and QB as their respective diameters. If a point S is such that the sum of the squares of the tangents drawn from S to these three circles is b², then the locus of S is

  1. x² + y² - 3ax + 2a² - b² = 0
  2. 3(x² + y²) - 9ax + 8a² - b² = 0
  3. x² + y² - 5ax + 6a² - b² = 0
  4. x² + y² - ax - b² = 0

Correct answer: 3(x² + y²) - 9ax + 8a² - b² = 0

Solution

With A(0,0),B(3a,0), P=(a,0),Q=(2a,0); the three circles have centres (a/2,0),(3a/2,0),(5a/2,0) and radius a/2. The sum of squared tangent lengths = sum of powers = 3(x^2+y^2) - 9ax + 8a^2. Setting this equal to b^2 gives 3(x^2+y^2) - 9ax + 8a^2 - b^2 = 0.

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