Question

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

• A. 4
• B. $-1$
• C. 0
• D. 1

Answer 1

The correct option is A

4

$S_1:x^2+y^2+2\lambda x+4=0$

$S_2:x^2+y^2-4\lambda y+8=0$

Since both represent real circles

$\therefore r_1\ge 0$ & $r_2\ge 0$

$\Rightarrow {\lambda }^2-4\ge 0\Rightarrow \lambda \le -2$ or $\lambda \ge 2$ ... (1)

And $4{\lambda }^2-8\ge 0\Rightarrow \lambda \le -\sqrt{2}$ or $\lambda \ge \sqrt{2}$ ... (2)

From equations $1$ & $2$

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

Among the given options, 4 lies in this range.