Question

If the point \( (2, \lambda) \) lies inside the circles \( x^2 + y^2 = 13 \) and \( x^2 + y^2 + x - 2y = 14 \), then \( \lambda \) lies in the set:

• A. \( (-\infty, -3) \cup (4, \infty) \)
• B. \( (-\infty, -1) \cup (3, \infty) \)
• C. \( [-3, 4] \)
• D. \( [-2, 3] \)

Hint:

Point Inside Circle}
Use condition: point is inside circle \( \Rightarrow \) expression<RHS
Reduce inequalities carefully and intersect domains
Check both geometric and algebraic bounds

Answer 1

The Correct Option is D

Let \( P = (2, \lambda) \) Circle 1: \( x^2 + y^2 = 13 \Rightarrow 2^2 + \lambda^2<13 \Rightarrow 4 + \lambda^2<13 \Rightarrow \lambda^2<9 \Rightarrow -3<\lambda<3 \) Circle 2: \( x^2 + y^2 + x - 2y = 14 \) Substitute \( x = 2 \), \( y = \lambda \): \[ 4 + \lambda^2 + 2 - 2\lambda<14 \Rightarrow \lambda^2 - 2\lambda<8 \Rightarrow \lambda^2 - 2\lambda - 8<0 \Rightarrow (\lambda - 4)(\lambda + 2)<0 \Rightarrow \lambda \in (-2, 4) \] Combine both conditions: \[ \lambda \in (-3, 3) \cap (-2, 4) = (-2, 3) \] Since boundaries are not strict, closed interval: \[ \lambda \in [-2, 3] \]