A) \[\frac{\sqrt{11}}{3}\]
B) \[\frac{\sqrt{13}}{3}\]
C) \[\frac{\sqrt{7}}{3}\]
D) \[\frac{1}{3}\]
Correct Answer: C
Solution :
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 \[=\sqrt{{{\left( -\frac{2}{3} \right)}^{2}}+0-\left( -\frac{1}{3} \right)}=\sqrt{\frac{4}{9}+\frac{1}{3}}=\frac{\sqrt{7}}{3}\]
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