The slopes of the focal chord of the parabola ${\mathrm{y}}^2=32\mathrm{x}$ which are tangents to the circle ${\mathrm{x}}^2+{\mathrm{y}}^2=4$ are

$$\begin{gathered} {\mathrm{y}}^2=32\mathrm{x} \\ 4\mathrm{a}=32 \\ \mathrm{a}=8 \end{gathered}$$

Focus=(8, 0)

Let 'm' be a slope of focal chord.

Equation chord is

 $$\begin{gathered} \mathrm{y}-0=\mathrm{m}\left(\mathrm{x}-8\right) \\ \mathrm{mx}-\mathrm{y}-8\mathrm{m}=0 \end{gathered}$$

it is  is a tangent to the circle ${\mathrm{x}}^2+{\mathrm{y}}^2=4$

$\mathrm{c}=\left(0, 0\right) \mathrm{r}=2$

$$\begin{gathered} \mathrm{r}=\mathrm{d} \\ 2=\frac{\left|-8\mathrm{m}\right|}{\sqrt{{\mathrm{m}}^2+1}} \\ 4\left({\mathrm{m}}^2+1\right)=64{\mathrm{m}}^2 \\ {\mathrm{m}}^2+1=16{\mathrm{m}}^2 \\ 15{\mathrm{m}}^2=1 \\ {\mathrm{m}}^2=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$15$}\right. \\ \mathrm{m}=\pm \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{15}$}\right. \end{gathered}$$