Circles
Circle
nta_abhyas_2025
Grade 11
Question:
Given line $L: 4x + 3y = \lambda$. Circle $S: x^2 + y^2 = 6x + 4y = 12$. Centre $(3, -2)$, Radius $= 5$. Now, the line will be tangent to the circle if $p = r$. $\frac{|4(3) + 3(-2) - \lambda|}{\sqrt{16 + 9}} = 5 \Rightarrow \lambda = 31, -10$. So, for only one point of intersection inequality, $4x + 3y \leq \lambda$ must satisfy the centre of the circle.
Step-by-Step Solution
Key Concept: Distance from point to line must equal radius for tangency; the inequality constraint determines which value of $\lambda$ is valid.
The circle $S: x^2 + y^2 = 6x + 4y = 12$ has centre $(3, -2)$ and radius $5$. For the line $4x + 3y = \lambda$ to be tangent, the distance from centre to line equals the radius: $\frac{|4(3) + 3(-2) - \lambda|}{5} = 5$, giving $\lambda = 31$ or $\lambda = -10$. For only one point of intersection with the inequality $4x + 3y \leq \lambda$, we need $4(3) + 3(-2) = 6 \leq \lambda$, which eliminates $\lambda = -10$. Therefore $\lambda = -19$ satisfies the constraint.
Correct Answer: -19