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

The correct option is D

$\frac{1}{3}$

Explanation for correct option

Step 1: Solve for the coefficients

Given equation is $\lambda x^2+(2\lambda -3)y^2-4x-1=0$

For the general equation of the circle is $x^2+y^2+2gx+2fy+c=0$ the radius is $\sqrt{g^2+f^2-c}$

If $a$ is the coefficient of $x^2$ and $b$ is the coefficient of $y^2$ then for circle $a=b$

$$\begin{gathered} \therefore a=b \\ \Rightarrow \lambda =2\lambda -3 \\ \Rightarrow \lambda =3 \end{gathered}$$

Therefore the equation of the circle can be represented as

$$\begin{gathered} 3x^2+3y^2-4x-1=0 \\ \Rightarrow x^2+y^2-\frac{4}{3}x-\frac{1}{3}=0 \end{gathered}$$

Now comparing with the general equation of the circle we get

$g=-\frac{2}{3},f=0,c=\frac{-1}{3}$

Step 2: Solve for the radius of the given circle

Radius $=\sqrt{g^2+f^2-c}$

$$\begin{gathered} =\sqrt{{\left(-\frac{2}{3}\right)}^2+{\left(0\right)}^2+\frac{1}{3}} \\ =\sqrt{\frac{7}{9}} \\ =\frac{\sqrt{7}}{3} \end{gathered}$$

Hence, Option (C) is correct