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
$ 3{{x}^{2}}+3{{y}^{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} $