$L\equiv y-mx-1=0$ of circle $S\equiv x^2+y^2-1=0$ touches the circle $S_1\equiv x^2+y^2-4x+1$.

Consider line, $L\equiv -mx+y-1=0$ and circle $S_1\equiv x^2+$ $y^2-4x+1$.

$\therefore$Centre of $S_1$ is $\left(2,0\right)$.

If the straight line touches the circle then,

$\sqrt{{\left(2\right)}^2-1}=\frac{\left|\left(-m\right)\left(2\right)+0-1\right|}{\sqrt{{\left(-m\right)}^2+1}}$

$\Rightarrow \sqrt{3}=\frac{\left|-2m-1\right|}{\sqrt{1+m^2}}$

$\Rightarrow {\left(\sqrt{3}\right)}^2=\frac{{\left(2m+1\right)}^2}{1+m^2}$

$\Rightarrow 3\left(1+m^2\right)=4m^2+1+4m$

$\Rightarrow m^2+4m-2=0$

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

$\Rightarrow m=\frac{-4\pm \sqrt{16+8}}{2}$

$\therefore m=-2\pm \sqrt{6}$

Now, $L=0$ is

$y-(-2\pm \sqrt{6})x-1=0$

$\Rightarrow y+(2\mp \sqrt{6})x-1=0 ....\left(\mathrm{i}\right)$

$\Rightarrow$ The chord of contact for a point $\left(h,k\right)$ with respect to $S$ is $ky+hx-1=0 ...\left(\mathrm{ii}\right)$

Comparing $\left(\mathrm{i}\right) \& \left(\mathrm{ii}\right)$, we get

$\left(h,k\right)\equiv \left(2\pm \sqrt{6},1\right)$