jmc

Substituting $y^2=8x$ into $x^2+y^2-2x-4y=0,$ we get $$x^2+6x-4y=0.$$Then $x^2+6x=4y.$ Squaring both sides, we get $$x^4+12x^3+36x^2=16y^2=128x.$$Hence, $$x^4+12x^3+36x^2-128x=0.$$We can take out a factor of $x,$ to get $$x(x^3+12x^2+36x-128)=0.$$We can check that $x=2$ is a root of the cubic, so we can also take out a factor of $x-2,$ to get $$x(x-2)(x^2+14x+64)=0.$$The quadratic factor has no real roots, so the real solutions are $x=0$ and $x=2.$

For $x=0,$ $y=0,$ and for $x=2,$ $y^2=16,$ so $y=\pm 4.$ We check that only $y=4$ satisfies the equation of the circle. Hence, the two intersection points are $(0,0)$ and $(2,4),$ and the distance between them is $\sqrt{2^2+4^2}=\sqrt{20}=\boxed{2\sqrt{5}}.$