Lengths  of  common  tangents  of  the  circles $x^2+y^2=6x, x^2+y^2+2x=0$  are

$C_1=\left(3,0\right),r_1=3$ 

$C_2=\left(-1,0\right),r_2=1$

$d=C_1{C}_2=4$

The length of transverse common tangent is $\sqrt{d^2-{(r_1+r_2)}^2} and the length of direct common tangent is \sqrt{d^2-{(r_1-r_2)}^2}$

$\sqrt{4^2-4^2},\sqrt{4^2-4}$

$=0,\sqrt{12}$

$=0,2\sqrt{3}$