Question

If the equation of a circle is $\lambda x^2+(2\lambda -3)y^2-4x+6y-1=0$ then the coordinates of centre are

• A. $(\frac{2}{3},1)$
• B. $(\frac{4}{3},-1)$
• C. $(\frac{2}{3},-1)$
• D. $(-\frac{2}{3},-1)$

Answer 1

The correct option is C $(\frac{2}{3},-1)$

$\because \lambda x^2+(2\lambda -3)y^2-4x+6y-1=0$

represents the equation of a circle

$\Rightarrow coeffof{x}^2=coeffof{y}^2$

$\Rightarrow \lambda =2\lambda -3$

$\Rightarrow \lambda =3$

$\therefore \lambda x^2+(2\lambda -3)y^2-4x+6y-1=0$

$\Rightarrow 3x^2+3y^2-4x+6y-1=0$

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

$g=-\frac{2}{3}andf=1$

$\therefore Centre=(\frac{2}{3},-1)$

Hence, option (b) is correct.