A common tangent to the circles $x^2+y^2=4$ and 

$(x-3{)}^2+y^2=1$ is

The centres of two circles are at $O (0, 0)$ and$(3, 0)$ and their radii are 2 and 1 unit respectively.

Clearly, $OC = 2 + 1$ i.e. distance between centres is equal to the sum of the radii of the circles. So, two circles touch each other externally and one of the common tangents (transverse) is given by $Si - Si = 0$i.e. $6x -12 = 0 or x = 2$. The point P of intersection of direct common tangents divides $OC$ externally in the ratio 2: 1. So, coordiantes of Pare $(6, 0)$. The equation of a line through $(6, 0)$ is

$y-0=m(x-6)$

or $mx-y-6m=0$ where mis the slope of the line

If it touches the circle $x^2+y^2=4$ then

$\left|\frac{0-0-6m}{\sqrt{m^2+1}}\right|=2\Rightarrow m^2+1=9m^2\Rightarrow m=\pm \frac{1}{2\sqrt{2}}$

Hence, the equations of direct common tangents are

$\pm \frac{x}{2\sqrt{2}}-y\mp \frac{3}{\sqrt{2}}=0\text{ or, }\pm x-2\sqrt{2}y\mp 6=0$