If the circles ( x + 1 ) ^ 2 + ( y + 2 ) ^ 2 = r ^ 2 and x ^ 2 + y...

Given,

${S}_{1} ≡ {\left(x + 1\right)}^{2} + {\left(y + 2\right)}^{2} = {r}^{2} and${S}_{2} ≡ {x}^{2} + {y}^{2} - 4x - 4y + 4 = 0 or {S}_{2} ≡ {\left(x - 2\right)}^{2} + {\left(y - 2\right)}^{2} = {2}^{2}$</p><p>Now, if two circles intersect at two distinct points then \left|{r}_{1} - {r}_{2}\right| < {C}_{1}{C}_{2} < {r}_{1} + {r}_{2}$.

$⇒ \left|r - 2\right| < \sqrt{{3}^{2} + {4}^{2}} < r + 2$

$⇒ \left|r - 2\right| < 5 < r + 2$

$⇒ \left|r - 2\right| < 5\text{and}r + 2 > 5$

$⇒ - 5 < r - 2 < 5\text{and}r > 3$

$⇒ - 3 < r < 7\text{and}r > 3$

$⇒ 3 < r < 7$