If the circles x ^ 2 + y ^ 2 - 16x - 20y + 164 = r ^ 2 and ( x - 4...

As we know, if two circles intersect each other, then

$\left|{r}_{1} - {r}_{2}\right| < {C}_{1}{C}_{2} < {r}_{1} + {r}_{2}$ ........(i)

Now for the first circle ${C}_{1}\left(8, 10\right)$ and ${r}_{1} = r$

For the second circle ${C}_{2}\left(4, 7\right)$ and ${r}_{2} = 6$

From (i)

\left|r - 6\right| < 5 < r + 6$</p><p>$⇒ \left|r - 6\right| < 5 . . . \left(ii\right)&5 < r + 6 . . . \left(iii\right)\left\{\left|r - 6\right| = \left\{$\begin{array}{ll}r - 6, r \ge 6 \\ 6 - r, r < 6\end{array}$\right\right\}

from \left(iii\right)$r > - 1 . . . \left(iv\right)$

from \left(ii\right)$</p><p>when$- 1 < r < 6$then$6 - r < 5 ⇒ 1 < r . . . \left(v\right)$</p><p>when$r \ge 6 ⇒ r - 6 < 5 ⇒ r < 11 . . . \left(vi\right)$</p><p>from \left(iv\right), \left(v\right), \left(vi\right)$ we get

$⇒ r \in \left(1, 11\right)$