If two circles $(x-1{)}^2+(y-3{)}^2=r^2$ and $x^2+y^2-8x+2y+8=0$ intersect in two distinct points, then 

If $d$is the distance between the centres of two circles of radii$r_1$ and $r_2$ , then they intersect in two distinct points iff

$\left|r_1-r_2\right|<d<r_1+r_2$

Here, radii of two circles are $r$ and 3 and distance between the centres is 5. So, the circles will intersect in two distinct points iff. 

$|r-3|<5<r+3\Rightarrow -2<r<8$ and $r>2\Rightarrow 2<r<8$