If $(-\frac{1}{3},-1)$ is a centre of similitude for the circles $x^2+y^2=1$ and $x^2+y^2-2x-6y-6=0$, then the length of common tangent of the circles is

The centers of the given circles are $C_1\left(0,0\right),C_2\left(1,3\right)$

Given that the Centre of similitude is $(-\frac{1}{3},-1)$

This point divides the line segment joining $C_1{C}_2$ in the ratio $1:4$ externally

Hence, this is external center of similitude. Direct common tangent will pass throug hte $\begin{array}{rcl}\sqrt{\frac{1}{9}+1+\frac{2}{3}+6-6}-\sqrt{\frac{1}{9}+1-1}& =& \frac{4}{3}-\frac{1}{3}\\ & =& 1\end{array}$

Therefore, the length of the direct common tangent is $1$ unit