If $\left(-\frac{1}{3}, -1\right)$ 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 circle is 
 

$$\begin{gathered} For given circles \\ {C}_1=(0,0),r_1=1 \\ and C_2=(1,3),r_2=4 \\ Now d=C_1{C}_2=\sqrt{10} \\ Given centre of similitude=(\frac{-1}{3},-1) \end{gathered}$$

Given centre of similitude is external centre of similitude 

Since it is external centre of Simultude then common tangent is direct  common tangent

$$\begin{gathered} Now length of direct common \mathrm{tangent}=\sqrt{{\mathrm{d}}^2-{({\mathrm{r}}_1-{\mathrm{r}}_2)}^2} \\ =1 \end{gathered}$$