JEE Main 2025 · 7 April · morning · Single correct · +4 / −1

Let the system of equations :
2 x+3 y+5 z=9 ; 7 x+3 y-2 z=8 ; 12 x+3 y-(4+λ) z=16-μ
have infinitely many solutions. Then the radius of the circle centred at (λ, μ) and touching the line 4 x=3 y is :

  1. (a) 7/5
  2. (b) 21/5
  3. (c) 7
  4. (d) 17/5
Answer and solution

Answer (a) 7/5

Solution

Maths sir · 5 steps

For infinitely many solutions, the third equation must be built from the first two, constant term included.

  1. The condition. The first two equations are independent. So for infinitely many solutions, the third equation must equal a (eq 1) + b (eq 2) exactly, including the right-hand side.
  2. Find a and b. Match the x and y coefficients: 2a + 7b = 12 and 3a + 3b = 3. The second gives a + b = 1, so a = -1 and b = 2. Check: 2(-1) + 7(2) = 12.
  3. Find λ and μ. For z: 5(-1) + (-2)(2) = -9 = -(4+λ), so λ = 5. This agrees with setting Δ = 15λ - 75 = 0. For the constants: 9(-1) + 8(2) = 7 = 16 - μ, so μ = 9.
  4. Distance to line. The circle touches 4x - 3y = 0, so its radius is the distance from (5, 9) to this line.

    r = (|4(5) - 3(9)|)/√(4² + 3²) = (|20 - 27|)/5 = 7/5

  5. So. The radius is 7/5, which is option (a).

The idea to keep. Infinitely many solutions means the third equation is a combination of the other two. Finding that combination gives both unknowns at once: one from the coefficients and one from the constants.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Determinants

Asked about 190 times in JEE Main since 2020. This question is from 2025.