The value of λ for which the set x , y : x 2 + y 2 - 6 x + 4 y ≤ 12 ∩ x , y : 4 x + 3 y ≤ λ contains only one point is

Step-by-step solution

Given line L : 4 x + 3 y = λ Circle S : x 2 + y 2 - 6 x + 4 y = 12 Centre 3 , – 2 , Radius = 5 Now, the line will be tangent to the circle ∴ p = r 4 3 + 3 - 2 - λ 5 = 5 ⇒ λ = 31 , – 19 So, for only one point of intersection inequality, 4 x + 3 y ≤ λ must not satisfy the centre of the circle ∴ λ = - 19