Let the set of all values of r, for which the circles $$ (x+1)^{2}+(y+4)^{2}=r^{2}$$ and $$ x^{2}+y^{2}-4x-2y-4=0$$ intersect at two distinct points be the interval $$( \alpha,\beta )$$. Then $$ \alpha\beta $$ is equal to

Given circles:
$$(x+1)^2+(y+4)^2=r^2$$

$$x^2+y^2-4x-2y-4=0;\Rightarrow;(x-2)^2+(y-1)^2=9$$

Centers:
$$C_1(-1,-4),\quad C_2(2,1),\quad r_2=3$$

Distance between centers:
$$d=\sqrt{(2+1)^2+(1+4)^2}=\sqrt{9+25}=\sqrt{34}$$

For two distinct intersections:
$$|r-3|<\sqrt{34}

$$\sqrt{34}-3

Thus interval ($$(\alpha,\beta)=(\sqrt{34}-3,;\sqrt{34}+3))$$
$$\alpha\beta=(\sqrt{34}-3)(\sqrt{34}+3)=34-9=25$$

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