Two common tangents to the circle $x^2+y^2=2 a^2$ and parabola $y^2=8 a x$ are
Two common tangents to the circle $x^2+y^2=2 a^2$ and parabola $y^2=8 a x$ are
$x=\pm(y+2 a)$
$y=\pm(x+2 a)$
$x=\pm(y+a)$
$y=\pm(x+a)$
Solution
Any tangent to the parabola $y^2=8 a x$ is
$
y=m x+\frac{2 a}{m}
$
If (i) is a tangent to the circle, $x^2+y^2=2 a^2$ then, $\quad \sqrt{2} a=\pm \frac{2 a}{m \sqrt{m^2+1}}$
$
\Rightarrow m^2\left(1+m^2\right)=2 \Rightarrow\left(m^2+2\right)\left(m^2-1\right)=0 ; \Rightarrow m=\pm 1
$
So, $y=\pm(x+2 a)$