$$
\begin{array}{ll}
& x^2+y^2-6 x-14 y+48=0 \\
\therefore \quad & \mathrm{C}_1(3,7), \mathrm{r}_1=\sqrt{10} \\
& \text { Again } x^2+y^2-6 x=0 \\
\therefore \quad & \mathrm{C}_2(3,0), \mathrm{r}_2=3
\end{array}
$$

Now $l\left(\mathrm{C}_1 \mathrm{C}_2\right)=$ distance between centres
$$
\begin{aligned}
& \therefore \quad l\left(\mathrm{C}_1 \mathrm{C}_2\right)=\sqrt{0^2+7^2}=7 \text { and } \\
& \\
& \mathrm{r}_1+\mathrm{r}_2=\sqrt{10}+3 < l\left(\mathrm{C}_1 \mathrm{C}_2\right)
\end{aligned}
$$
$\Rightarrow$ The given circles are disjoint.
$\Rightarrow$ Number of common tangents is 4 .