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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Circle

JEE Main 2026 · 21 January · Morning Shift · Mathematics · Circle

Let PQ and MN be two straight lines touching the circle x^ 2 +y^ 2...

Let PQ and MN be two straight lines touching the circle x2+y2−4x−6y−3=0x^{2}+y^{2}-4 x-6 y-3=0 at the points A and B respectively. Let O be the centre of the circle and ∠AOB=π/3\angle \mathrm{AOB}=\pi / 3. Then the locus of the point of intersection of the lines PQ and MN is :

Options

  1. Ax2+y2−12x−18y−25=0x^{2}+y^{2}-12 x-18 y-25=0
  2. B3(x2+y2)−18x−12y+25=03\left(x^{2}+y^{2}\right)-18 x-12 y+25=0
  3. C3(x2+y2)−12x−18y−25=03\left(x^{2}+y^{2}\right)-12 x-18 y-25=0
  4. Dx2+y2−18x−12y−25=0x^{2}+y^{2}-18 x-12 y-25=0

Correct answer

C. 3(x2+y2)−12x−18y−25=03\left(x^{2}+y^{2}\right)-12 x-18 y-25=0

Step-by-step solution

Circle: (x−2)2+(y−3)2=16(x-2)^2 + (y-3)^2 = 16, center O = (2, 3), radius r=4r = 4.
For tangents from external point P touching at A and B with ∠AOB=π/3\angle AOB = \pi/3:
In quadrilateral OAPB: ∠APB=π−π/3=2π/3\angle APB = \pi - \pi/3 = 2\pi/3, so ∠OPA=π/3\angle OPA = \pi/3.
In right triangle OAP: sin⁡(π/3)=4OP\sin(\pi/3) = \frac{4}{OP}.
OP=83=833OP = \frac{8}{\sqrt{3}} = \frac{8\sqrt{3}}{3}.
Locus: (x−2)2+(y−3)2=643(x-2)^2 + (y-3)^2 = \frac{64}{3}.
3(x2+y2)−12x−18y−25=03(x^2 + y^2) - 12x - 18y - 25 = 0.
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