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AP EAMCET202120 Aug 2021Morning ShiftMathematicsStraight LinesActual

The locus of a point which is at a distance of 4 units from ( 3 , - 2 ) in x y -plane is

Options

  1. Ax 2 + y 2 + 6 x - 4 y + 16 = 0
  2. Bx 2 + y 2 - 6 x - 4 y + 3 = 0
  3. Cx 2 + y 2 - 6 x + 4 y - 16 = 0
  4. Dx 2 + y 2 - 6 x + 4 y - 3 = 0

Correct answer

D. x 2 + y 2 - 6 x + 4 y - 3 = 0

Step-by-step solution

Let the point be h , k Distance between the two points will be = h - 3 2 + k + 2 2 ⇒ 4 = h - 3 2 + k + 2 2 Squaring both sides we get, 16 = h 2 + 9 - 6 h + k 2 + 4 k + 4 ⇒ h 2 - 6 h + k 2 + 4 k - 3 = 0 Therefore, required locus is x 2 + y 2 - 6 x + 4 y - 3 = 0

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