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−2…

Given circles:

\({(x−2)}^{2}+{(y−1)}^{2}={3}^{2}\\ \text{and}{(x+1)}^{2}+{(y+4)}^{2}={r}^{2}\)

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}}

\(\Rightarrow \left|r−3\right|<\sqrt{34}\text{  }&\text{  }r+3>\sqrt{34}\)

\(\Rightarrow −\sqrt{34}\sqrt{34}−3\)

i.e. \(r=\left(3−\sqrt{34},\text{  }3+\sqrt{34}\right)\cap \left(\sqrt{34}−3,∞\right)\)

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

\(∴\alpha \beta =\left(\sqrt{34}−3\right)\left(\sqrt{34}+3\right)\)

\(=34–9=25\)