Question

If $3x^2+2\lambda xy+3y^2+(6-\lambda )x+(2\lambda -6)y-21=0$ is the equation of circle then its radius is

• A. $1$
• B. $3$
• C. $2\sqrt{2}$
• D. $4$

Answer 1

The correct option is B $3$

$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ is represent equation of circle if
$a=b$ and $h=0$
So, $3x^2+2\lambda xy+3y^2+(6-\lambda )x+(2\lambda -6)y-21=0$
is an equation of the circle
If, $\lambda =0$
$\therefore x^2+y^2+\frac{6}{3}x-\frac{6}{3}y-\frac{21}{3}=0$
$x^2+y^2+2x-2y-7=0$
$\Rightarrow radius=\sqrt{1^2+1^2+7}$
$=3.$