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JEE Main202621 January 2026Morning ShiftMathematicsCircleActual

Let PQ and MN be two straight lines touching the circle x²+y²-4 x-6 y-3=0 at the points A and B respectively. Let O be the centre of the circle and AOB = / 3 . Then the locus of the point of intersection of the lines PQ and MN is :

Options

  1. Ax²+y²-12 x-18 y-25=0
  2. B3 (x²+y² )-18 x-12 y+25=0
  3. C3 (x²+y² )-12 x-18 y-25=0
  4. Dx²+y²-18 x-12 y-25=0

Correct answer

C. 3 (x²+y² )-12 x-18 y-25=0

Step-by-step solution

Circle: (x-2)^2 + (y-3)^2 = 16 , center O = (2, 3), radius r = 4 . For tangents from external point P touching at A and B with AOB = /3 : In quadrilateral OAPB: APB = - /3 = 2 /3 , so OPA = /3 . In right triangle OAP: ( /3) = 4 OP . OP = 8 3 = 8 3 3 . Locus: (x-2)^2 + (y-3)^2 = 64 3 . 3(x^2 + y^2) - 12x - 18y - 25 = 0 .

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