If $y=f\left(x\right)=ax+b$  is a tangent to circle $x^2+y^2+2x+2y-2=0$  then the value of  ${(a+b)}^2+2(a-b)(a+b+1)$ is equal to

Substitute y in equation of circle and put  $△=0$
$x^2+{(ax+b)}^2+2x+2(ax+b)-2=0$
$x^2(1+a^2)+x(2ab+2+2a)+(b^2+2b-2)=0$
${(a+1+ab)}^2=(1+a^2)(b^2+2b-2)$
Hence  ${(a+b)}^2+2(a-b)(a+b+1)=0$