Show that the equation of the common tangent to the circle x²+y² = 2a² the parabola y²= 8ax is y =$\pm$ (x+2a)

Given the equation of the parabola is y² = 8ax........(1)
and the circle is ${\mathrm{x}}^2+{\mathrm{y}}^2=2{\mathrm{a}}^2$.......(2)
Any tangent of (1) is $\mathrm{y}=\mathrm{mx}+\frac{2\mathrm{a}}{\mathrm{m}}\Rightarrow \mathrm{mx}-\mathrm{y}+\frac{2\mathrm{a}}{\mathrm{m}}=0$......(3)
line (3) is also a tangent to circle (2) $\Rightarrow \frac{\left|\frac{2\mathrm{a}}{\mathrm{m}}\right|}{\sqrt{\left(1+{\mathrm{m}}^2\right)}}=\mathrm{a}\sqrt{2}$
If $\frac{4}{{\mathrm{m}}^2}=2\left(1+{\mathrm{m}}^2\right)\text{ i.e., }{\mathrm{m}}^4+{\mathrm{m}}^2-2=0$
If $\left({\mathrm{m}}^2+2\right)\left({\mathrm{m}}^2-1\right)=0\text{ i.e., }\Rightarrow {\mathrm{m}}^2\ne -2,{\mathrm{m}}^2=1\Rightarrow \mathrm{m}\pm 1$
$\therefore$The equation of the common tangent (1) & (2) are
$\mathrm{y}=\pm \mathrm{x}\pm 2\mathrm{a}\Rightarrow \mathrm{y}=\pm (\mathrm{x}+2\mathrm{a})$