$A$ tangent $PT$is drawn to the circle $x^2+y^2=4$ at the point $P(\sqrt{3},1).$ A straight line$L,$perpendicular to $PT$ is a tangent to the circle

$(x-3{)}^2+y^2=1.$ the possible equation of $L$ is

The equation of tangent at $P(\sqrt{3},1)$ to the circle $x^2+y^2=4\text{ is }\sqrt{3}x+y=4$ 

Its slope is $-\sqrt{3}$ So, slope of a line perpendicular to$PT$ is $\frac{1}{\sqrt{3}}$

The equation of tangents of slope $\frac{1}{\sqrt{3}}$ to the circle

$(x-3{)}^2+(y-0{)}^2=1^2$ are 

$y-0=\frac{1}{\sqrt{3}}(x-3)\pm \sqrt{1+{\left(\frac{1}{\sqrt{3}}\right)}^2}\text{ or, }\sqrt{3}y=x-3\pm 2$

or $x-\sqrt{3}y-1=0\text{ and }x-\sqrt{3}y-5=0$