Question

If the equation $\lambda x^2+(2\lambda -3)y^2-4x-1=0$ represents a circle, then its radius is

• A. $\frac{\sqrt{11}}{3}$
• B. $\frac{\sqrt{13}}{3}$
• C. $\frac{\sqrt{7}}{3}$
• D. $\frac{1}{3}$

Answer 1

Answer: (C)

Given, equation is $\lambda x^2+(2\lambda -3)y^2-4x-1=0$
Here, $a=\lambda ,b=2$
It represents a circle, if $a=b$
$\Rightarrow$ $\lambda =2\lambda -3$
$\Rightarrow$ $\lambda =3$
Also, $h=0$
Then, equation becomes
$3x^2+3y^2-4x-1=0$
$\Rightarrow$ $x^2+y^2-\frac{4}{3}x-\frac{1}{3}=0$
Here, $g=-\frac{2}{3},c=-\frac{1}{3},f=0$
$\therefore$ Radius
$=\sqrt{{\left(-\frac{2}{3}\right)}^2+0-\left(-\frac{1}{3}\right)}=\sqrt{\frac{4}{9}+\frac{1}{3}}=\frac{\sqrt{7}}{3}$