Circles Part 2

Concept:Two circles intersect at two distinct points if the distance between their centers is between the absolute difference and the sum of their radii.Explanation:Circle 1: $(x+1)^2+(y+4)^2=r^2$ gives center $C_1=(-1,-4)$ and radius $r_1=r$.Circle 2: $x^2+y^2-4x-2y-4=0$ can be rewritten as $(x-2)^2+(y-1)^2=9$. So center $C_2=(2,1)$ and radius $r_2=3$.Distance $C_1C_2=\sqrt{(2+1)^2+(1+4)^2}=\sqrt{9+25}=\sqrt{34}$.Condition for two intersection points: $|r_1-r_2| < C_1C_2 < r_1+r_2$.Thus $|r-3| < \sqrt{34} < r+3$.From $\sqrt{34} < r+3$ we get $r > \sqrt{34}-3$.From $|r-3| < \sqrt{34}$ we get $3-\sqrt{34} < r < 3+\sqrt{34}$. Since $3-\sqrt{34}$ is negative, the positive lower bound is $\sqrt{34}-3$.So $r$ lies in $(\sqrt{34}-3,\ \sqrt{34}+3)$. Hence $\alpha = \sqrt{34}-3$ and $\beta = \sqrt{34}+3$.Then $\alpha\beta = (\sqrt{34}-3)(\sqrt{34}+3)=34-9=25$.Answer:$\alpha\beta = 25$, which corresponds to option D.