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

$x^2+y^2=6x\Rightarrow x^2+y^2-6x=0$

  $C_1=(3,0),r_1=\sqrt{9+0-0}=3$

  $x^2+y^2+2x=0$

 $C_2=(-1,0)=r_2=1$

 $\begin{array}{l}{C}_1{C}_2=\sqrt{{(3+1)}^2+0}=4\\ \therefore C_1{C}_2=r_1+r_2\end{array}$

 $\therefore$ There exists 3 tangents

   Length of the transverse common tangent is 0

  Length of the direct common tangent is $\sqrt{d^2-{(r_1-r_2)}^2}$

   $\begin{array}{l}=\sqrt{4^2-{(3-1)}^2}\\ =\sqrt{16-4}\\ =\sqrt{12}=2\sqrt{3}\end{array}$