Question

If the equation $3x^2+3y^2+6\lambda x+2\lambda =0$ represents a circle, then the value of $\lambda$ lies in

• A. $(-\infty ,0)\cup (\frac{2}{3},\infty )$
• B. $(-\infty ,-1)\cup [\frac{1}{3},\infty )$
• C. $(-\infty ,0]\cup [\frac{2}{3},\infty )$
• D. none of these

Answer 1

Answer: (C)

$(-\infty ,0]\cup [\frac{2}{3},\infty )$

First, we will represent this equation in center-radius form,that is $(x-h{)}^2+(y-k{)}^2=r^2$

Add and subtract $3{\lambda }^2$ on the $LHS$ of the equation.

$\Rightarrow$ $3x^2+3y^2+6\lambda x+2\lambda -3{\lambda }^2+3{\lambda }^2=0$

$\Rightarrow$ $3(x^2+2\lambda x+{\lambda }^2+y^2)+2\lambda -3{\lambda }^2=0$

$\Rightarrow$ $3\left(\right(x+\lambda {)}^2+y^2)+2\lambda -3{\lambda }^2=0$

$\Rightarrow$$(x+\lambda {)}^2+y^2=\frac{3{\lambda }^2-2\lambda }{3}$

This equation will represent circle if $\frac{3{\lambda }^2-2\lambda }{3}\ge 0$($\because r\ge 0$ for a circle)

$\Rightarrow$$3{\lambda }^2-2\lambda \ge 0$

$\Rightarrow$$\lambda (3\lambda -2)\ge 0$

This means either $\lambda \ge \frac{2}{3}$ or $\lambda \le 0$

$\therefore \lambda ϵ(-\infty ,0]\cup [\frac{2}{3},\infty )$

Hence, $\left(C\right)$