If $-3{\mathrm{l}}^2-6\mathrm{l}+6{\mathrm{m}}^2-1=0$ then equation of the circle for which $\mathrm{lx}+\mathrm{my}+1=0$ is a tangent

$\begin{array}{l}6{\mathrm{l}}^2+6{\mathrm{m}}^2=9{\mathrm{l}}^2+6\mathrm{l}+1\\ \Rightarrow \left(\frac{3\mathrm{l}+1}{\sqrt{{\mathrm{l}}^2+{\mathrm{m}}^2}}\right)=\sqrt{6}\end{array}$

perpendicular distance from (3, 0) to the line $\mathrm{lx}+\mathrm{my}+1=0$ is $\sqrt{6}$

$\therefore lx+my+1=0 is \tan gent to the circle with centre (3,0)$  and $\sqrt{6}$ is radius, 

Required circle is $(\mathrm{x}-3{)}^2+{\mathrm{y}}^2=6$