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JEE Advanced2026MathematicsCircleActual

Match each entry in List-I to the correct entry in List-II and choose the correct option. List-I List-II (P) The circle with centre (1, 2) and touching the straight line 3x + 4y = 1 , passes through (1) the point (1, 1) (Q) The common tangent to the circle x^2 + y^2 = 2 and the parabola y^2 = 8x with positive slope, passes through (2) the point (7, 9) (R) Let M be the end point of the latus rectum of the ellipse 3x^2

Options

  1. A(P) (3), (Q) (4), (R) (1), (S) (2)
  2. B(P) (3), (Q) (2), (R) (1), (S) (5)
  3. C(P) (3), (Q) (2), (R) (4), (S) (5)
  4. D(P) (4), (Q) (1), (R) (2), (S) (3)

Correct answer

B. (P) (3), (Q) (2), (R) (1), (S) (5)

Step-by-step solution

For (P): The radius of the circle is the perpendicular distance from (1, 2) to 3x + 4y - 1 = 0 . r = |3(1) + 4(2) - 1| 3^2 + 4^2 = 10 5 = 2 The equation of the circle is (x - 1)^2 + (y - 2)^2 = 4 . Checking the given points, (3, 2) satisfies the equation. Thus, (P) (3). For (Q): The equation of a tangent to y^2 = 8x with slope m is y = mx + 2 m . Since it is also a tangent to x^2 + y^2 = 2 , the perpendicular distance from (0, 0) to the line mx - y + 2 m = 0 is equal to the radius 2 . | 2 m | m^2 + 1 = 2 4 m^2 = 2(

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