Any tangent to parabola $y^{2}=8 x$ is $y=m x+\frac{2}{m} \ldots$ (i) It touches the circle $x^{2}+y^{2}-12 x+4=0$, if the length of perpendicular from the centre $(6,0)$ is equal to radius $\sqrt{32}$. $\therefore \frac{6 m +\frac{2}{ m }}{\sqrt{ m ^{2}+1}}=\pm \sqrt{32} $ $\Rightarrow\left(3 m +\frac{1}{ m }\right)^{2}=8\left( m ^{2}+1\right)$ $\Rightarrow\left(3 m ^{2}+1\right)^{2}=8\left( m ^{4}+ m ^{2}\right)$ $\Rightarrow m ^{4}-2 m ^{2}+1=0 $ $\Rightarrow m =\pm 1$ Hence, the required tangents are $y=x+2$ and $y=-x-2$.