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MathematicsList match+3 / -1Question 15 of 51

JEE Advanced 2024 · Paper 1, Mathematics question 15

Let the straight line y=2xy = 2x touch a circle with center (0,α)(0, \alpha), α>0\alpha > 0, and radius rr at a point A1A_1. Let B1B_1 be the point on the circle such that the line segment A1B1A_1B_1 is a diameter of the circle. Let α+r=5+5\alpha + r = 5 + \sqrt{5}.

Match each entry in List-I to the correct entry in List-II.

The correct option is

List-I

  • P
    α\alpha equals
  • Q
    rr equals
  • R
    A1A_1 equals
  • S
    B1B_1 equals

List-II

  • 1
    (−2,4)(-2, 4)
  • 2
    5\sqrt{5}
  • 3
    (−2,6)(-2, 6)
  • 4
    55
  • 5
    (2,4)(2, 4)
Choose one option
Answer and worked solution

Official answer: C

Tangency. The distance from (0,α)(0,\alpha) to 2x−y=02x - y = 0 equals the radius: r=α5r = \dfrac{\alpha}{\sqrt5}. Then
α+r=α 5+15=5+5=5(5+1)  ⟹  α=5, r=5.\alpha + r = \alpha\,\frac{\sqrt5+1}{\sqrt5} = 5 + \sqrt5 = \sqrt5(\sqrt5+1) \implies \alpha = 5,\ r = \sqrt5.

A1A_1 is the foot of the perpendicular from (0,5)(0,5) to 2x−y=02x - y = 0:
(0,5)−2⋅0−55(2,−1)=(2,4).(0,5) - \frac{2\cdot0 - 5}{5}(2,-1) = (2,4).

B1B_1 =2(0,5)−(2,4)=(−2,6)= 2(0,5) - (2,4) = (-2,6).

So (P)→(4), (Q)→(2), (R)→(5), (S)→(3): option (C).

Topics:Tangents and normalsDistance of a point from a line
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