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)

$\left|r-6\right|<5<r+6$

$\Rightarrow \left|r-6\right|<5 ...\left(\mathrm{ii}\right)$ & $5<r+6 ...\left(\mathrm{iii}\right)$  $\left\{\left|r-6\right|=\left\{\begin{array}{l}r-6, r\ge 6\\ 6-r, r<6\end{array}\right.\right\}$

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

from $\left(\mathrm{ii}\right)$ 

when $-1<r<6$ then $6-r<5\Rightarrow 1<r ...\left(\mathrm{v}\right)$

when $r\ge 6\Rightarrow r-6<5\Rightarrow r<11 ...\left(\mathrm{vi}\right)$

from $\left(\mathrm{iv}\right), \left(\mathrm{v}\right), \left(\mathrm{vi}\right)$ we get

$\Rightarrow r\in \left(1, 11\right)$