End points of the latus rectum of the parabola $y^2=4x$ and $\left(1,\pm 2\right).$
Equation of the normals at points $\left(1,\pm 2\right)$ are
$y=-x+3$ and $y=x-3$
rr $x+y-3=0$ and $x-y-3=0$
These lines are tangent to the circle ${\left(x-3\right)}^2+{\left(y+2\right)}^2=r^2$
$\therefore \left|\frac{3\pm 2-3}{\sqrt{1+1}}\right|=r=\sqrt{2}$ 

Hence, $r^4=4$