If the line $x-1=0$  is the directrix of the parabola  $y^2-kx+8=0,k\ne 0$ and the parabola intersects the circle $x^2+y^2=4$  in two real distinct points, then the value of  $\left|k\right|$ is

The equation of the parabola can be written as  $y^2=k(x-\frac{8}{k})$
Equation of the directrix is  $x=h-a$
$\begin{array}{l}1=\frac{8}{k}-\frac{k}{4}(\because x=1)\\ \Rightarrow k^2+4k-32=0\Rightarrow k=-8or4\end{array}$
For $k=4$ , the parabola is $y^2=4(x-2)$ whose vertex is (2,0) and touches the circle $x^2+y^2=4$ at the vertex
For $k=-8$ , the parabola is $y^2=-8(x+1).$ Which intersects the circle $x^2+y^2=4$ at two real distinct points as the vertex (-1,0) of the parabola lies inside the circle.