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