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

If the point $(2, \lambda)$ lies inside the circles $x^2+y^2=13$ and $x^2+y^2+x-2 y=14, \lambda$ lies in the set
  1. $(-\infty,-3) \cup(4, \infty)$
  2. $(-\infty,-1) \cup(3, \infty)$
  3. $[-3,4]$
  4. $(-2,3)$

Solution

Given, $S_1 \equiv x^2+y^2-13, S_2 \equiv x^2+y^2+x-2 y-14$ According to question, $S_1(2, \lambda) < 0$ $\Rightarrow 2^2+\lambda^2-13 < 0 \Rightarrow \lambda^2 < 3^2$ $\Rightarrow \quad-3 < \lambda < 3$ ...(A) and $S_2(2, \lambda) < 0$ $\begin{aligned} & \Rightarrow \quad 2^2+\lambda^2+2-2 \lambda-14 < 0 \Rightarrow \lambda^2-2 \lambda-8 < 0 \\ & \Rightarrow \quad(\lambda-4)(\lambda+2) < 0\end{aligned}$ $-2 < \lambda < 4$ ...(B)
$\therefore \quad \lambda \in(-2,3)$

Asked in: AP EAMCET 2022 (07 Jul Shift 1)

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