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ExamsJEE MainMaths

A circle goes through the point (a,b) and intersects the circle x²+y²=4 at right angles. The set of all possible centres of such circles is

  1. 2ax - 2by - (a² + b² + 4) = 0
  2. 2ax + 2by - (a² + b² + 4) = 0
  3. 2ax - 2by + (a² + b² + 4) = 0
  4. 2ax + 2by + (a² + b² + 4) = 0

Correct answer: 2ax + 2by - (a² + b² + 4) = 0

Solution

Let the circle be x^2+y^2+2gx+2fy+c=0. Orthogonality with x^2+y^2=4 gives c=4. Passing through (a,b): a^2+b^2+2ga+2fb+4=0. With center (x,y)=(-g,-f), the locus of centers is 2ax+2by-(a^2+b^2+4)=0.

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