Question

The equation ${\lambda }^2{x}^2+({\lambda }^2-5\lambda +4)xy+(3\lambda -$ $2)y^2-8X+12y-4=0$ will represent a circle, if $\lambda =$

• A. $1$
• B. $4$
• C. $2$
• D. $3$

Answer 1

The correct option is A $1$

${\lambda }^2{x}^2+({\lambda }^2-5\lambda +4)xy+(3\lambda -2)y^2-8x+12y-4=0$
This equation represents a circle.
As we know, $a{x}^2+2hxy+b{y}^2+2gn+2fy+c=0$ represent a equation of circle if
$a=b$ and $h=0$.
So, ${\lambda }^2=3\lambda -2$ and ${\lambda }^2-5\lambda +4=0$
${\lambda }^2-3\lambda +2=0$ and $\lambda =\frac{5\pm \sqrt{9}}{2}$
$\lambda =\frac{3\pm \sqrt{1}}{2}$ and $\lambda =4,1$
$\lambda =2,1$ and $\lambda =4,1$
So, common $\lambda =1$