The equations of the tangents to the circle x 2 + y 2 = 4 drawn from the point ( 4 , 0 ) are

A line passing through point $\left(4, 0\right)$, with slope $m$ is given by

$y=mx-4m ...\left(1\right)$

$\Rightarrow mx-y-4m=0$

Given, $x^2+y^2=4$, Centre $\left(0, 0\right)$ and Radius$=2$

As, the given line is tangent to the circle, hence, the perpendicular distance from the centre to the tangent is equal to the radius of the given circle.

$\left|\frac{-4m}{\sqrt{1+m^2}}\right|=2 \Rightarrow {\left(\frac{-4m}{\sqrt{1+m^2}}\right)}^2=4\Rightarrow \frac{16m^2}{1+m^2}=4\Rightarrow m=\pm \frac{1}{\sqrt{3}}$

So, the equation of the tangent is $y=\pm \frac{1}{\sqrt{3}}\left(x-4\right)$