If $a>2b>0$  then the positive value of $m$  for which $y=mx-b\sqrt{1+m^2}$  is a common tangent to $x^2+y^2=b^2$ and ${\left(x-a\right)}^2+y^2=b^2$ is

$y=mx-b\sqrt{1+m^2}$ is a tangent to the circle $x^2+y^2=b^2$ for all values of m. If it also touches the circle ${\left(x-a\right)}^2+y^2=b^2$ , then the length of the perpendicular from its centre $\left(a,0\right)$  on this line is equal to the radius b of the circle, which gives

$\frac{ma-b\sqrt{1+m^2}}{\sqrt{1+m^2}}=\pm b$

Taking negative value on R.H.S. We get $m=0,$ so we neglect it.

Taking the positive value on R.H.S. we get

$ma=2b\sqrt{1+m^2}$

$\Rightarrow m^2\left(a^2-4b^2\right)=4b^2$

$\Rightarrow m=\frac{2b}{\sqrt{a^2-4b^2}}$