The lengths of the common tangents of the circles $x^2+y^2+4x=0$ and $x^2+y^2-6x=0$ is

$$\begin{gathered} {\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}=0 \\ C_1=(-2, 0) {\mathrm{r}}_1=2 \\ {\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}=0 \\ C_2=(3, 0),r_2=3 \end{gathered}$$

$Now d=C_1{C}_2=5$

Length of direct common tangent

$$\begin{gathered} =\sqrt{{\mathrm{d}}^2-{({\mathrm{r}}_1-{\mathrm{r}}_2)}^2} \\ =\sqrt{25-1} \\ =\sqrt{24} \\ =2\sqrt{6} \\ length of transverse common \tan gent=\sqrt{{\mathrm{d}}^2-{({\mathrm{r}}_1+{\mathrm{r}}_2)}^2}=0 \end{gathered}$$

length of Common tangents are $2\sqrt{6}, 0$