Circles

Given circles:

$$\begin{gathered} {(x-2)}^2+{(y-1)}^2=3^2 \\ \text{and} {(x+1)}^2+{(y+4)}^2=r^2 \end{gathered}$$

Since the circles intersect each other at two different points.

So, $\left|r_1-r_2\right|<c_1{c}_2<r_1+r_2$

$\Rightarrow \left|r-3\right|<\sqrt{{(2+1)}^2+{(1+4)}^2}<r+3$

$\Rightarrow \left|r-3\right|<\sqrt{34}\&r+3>\sqrt{34}$

$\Rightarrow -\sqrt{34}<r-3<\sqrt{34}\&r>\sqrt{34}-3$

i.e. $r=\left(3-\sqrt{34},3+\sqrt{34}\right)\cap \left(\sqrt{34}-3,\infty \right)$

i.e. $r\in \left(\sqrt{34}-3,\sqrt{34}+3\right)$

$\therefore \alpha \beta =\left(\sqrt{34}-3\right)\left(\sqrt{34}+3\right)$

$= 34 – 9=25$