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JEE Main202225 Jun 2022Morning ShiftMathematicsCircleActual

Let a circle C touch the lines L 1 : 4 x - 3 y + K 1 = 0 and L 2 : 4 x - 3 y + K 2 = 0 , K 1 , K 2 ∈ R . If a line passing through the centre of the circle C intersects L 1 at - 1 , 2 and L 2 at 3 , - 6 , then the equation of the circle C is

Options

  1. Ax - 1 2 + y - 2 2 = 4
  2. Bx - 1 2 + y + 2 2 = 16
  3. Cx + 1 2 + y - 2 2 = 4
  4. Dx - 1 2 + y - 2 2 = 16

Correct answer

B. x - 1 2 + y + 2 2 = 16

Step-by-step solution

Given, L 1 = 4 x - 3 y + K 1 = 0 ;   L 2 = 4 x - 3 y + K 2 = 0 Here line L 1 and L 2 are parallel. So, Now point A - 1 , 2 will satisfy L 1 so, - 4 - 3 × 2 + K 2 = 0 ⇒   K 2 = 10 Also point B 3 , - 6 will satisfy L 2 = 4 x - 3 y + K 2 So 4 × 3 - 3 × - 6 + K 2 = 0     ⇒ K 2 = - 30 Now distance between the parallel line will be the diameter Diameter = K 1 - K 2 4 2 + 3 2 = 10 + 30 5 = 8 So radius 8 2 = 4 Now mid-point of A B will give us centre of circle by symmetry, so

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