Question

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

• A. $1$
• B. $\sqrt{2}$
• C. $2$
• D. $5$

Answer 1

The correct option is D $5$

The general equation of a circle is,

$x^2+y^2+2gx+2fy+c=0$

Whose radius is $r=\sqrt{g^2+f^2-c}$

$\therefore$ For a real circle $g^2+f^2-c>0$

Now, We have two circle,

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

the radius of this circle $=r_1=\sqrt{{\lambda }^2-4}$

for a real circle ${\lambda }^2-4>0$

$\Rightarrow {\lambda }^2>4$

$\Rightarrow \lambda >2$ or $\lambda <-2⟶\left(1\right)$

also,

$C2:x^2+y^2-4\lambda y+8=0$

the radius of this circle $=r_2=\sqrt{{4\lambda }^2-8}$

For a real circle ${4\lambda }^2-8>0$

$\Rightarrow {\lambda }^2>2$

$\Rightarrow \lambda >\sqrt{2}$ or $\lambda <-\sqrt{2}⟶\left(2\right)$

Combining $\left(1\right)$ and $\left(2\right)$ we have

$\lambda ϵ(-\infty ,-2)\cup (2,\infty )$

From the options,

$\therefore$ the value of $\lambda$ can be $3$ or $5$

Hence correct options are $C$ & $D$.