If the equation $x^2+2\sqrt{2}xy+2y^2+4x+4\sqrt{2}y+1=0$ represents tangents to same circle then circumference of circle is

Given Equation $x^2+2\sqrt{2}xy+2y^2+4x+4\sqrt{2}y+1=0--\left(1\right)$

Comparing with  $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$

$\begin{array}{l}a=1,h=\sqrt{2},b=2,g=2,f=2\sqrt{2},c=1\\ Now{h}^2=abanda{f}^2=b{g}^2\end{array}$

(1) represent pair of parallel lines

Diameter of circle

$\begin{array}{l}=2\sqrt{\frac{g^2-ac}{a\left(a+b\right)}}=2\sqrt{\frac{4-1}{1\left(1+2\right)}}=2\left(1\right)=2\\ Radiusofcircle=\frac{1}{2}\left(diameter\right)\\ Circumference=2\pi r=2\pi \end{array}$