Question

If the equation $x^2+y^2+2\lambda x+4=0$ and $x^2+y^2+4\lambda y+8=0$ represent real circles then the value of $\lambda$ can be

• A. $5$
• B. $2$
• C. $3$
• D. $Allofthese$

Answer 1

The correct option is D $Allofthese$

Solution:- (D) all of these

$C_1:x^2+y^2+2\lambda x+4=0$

$\Rightarrow x^2+2\lambda x+{\lambda }^2-{\lambda }^2+y^2+4=0$

$\Rightarrow {(x+\lambda )}^2+y^2-{\lambda }^2+4=0$

$\Rightarrow {(x+\lambda )}^2+y^2={\lambda }^2-4$

${r_1}^2={\lambda }^2-4\ge 0$

$\Rightarrow (\lambda -2)(\lambda +2)\ge 0$

$\Rightarrow \lambda \in (-\infty ,-2]\cup [2,\infty ).....\left(1\right)$

$C_2:x^2+y^2+4\lambda y+8=0$

$\Rightarrow x^2+y^2+4\lambda y+4{\lambda }^2-4{\lambda }^2+8=0$

$\Rightarrow x^2+{(y+2\lambda )}^2-4{\lambda }^2+8=0$

$\Rightarrow x^2+{(y+2\lambda )}^2=4{\lambda }^2-8$

${r_2}^2={4\lambda }^2-8\ge 0$

$\Rightarrow {\lambda }^2-2\ge 0$

$\Rightarrow \lambda \in (-\infty ,-\sqrt{2}]\cup [\sqrt{2},\infty ).....\left(2\right)$

From ${eq}^n\left(1\right)\&\left(2\right)$, we get

$\lambda \in \left(1\right)\cap \left(2\right)$

$\Rightarrow \lambda \in (-\infty ,-2]\cup [2,\infty )$

Since all of the given values lie within the range.

Hence the correct answer is (D) all of these.