Let $C_1=$ Centre of first circle
$$
=\left[-\frac{1}{2} \times(-2),-\frac{1}{2} \times(-6)\right]=(1,3)
$$
$C_2=$ Centre of second circle
$$
=\left[-\frac{1}{2} \times(-8),-\frac{1}{2} \times(2)\right]=(4,-1)
$$
$r_1=$ radius of first circle $=\sqrt{1+9-10+r^2}=|r|$
$r_2=$ radius of second circle $=\sqrt{16+1-8}=3$
According to the question, $\left|r_2-r_1\right| < c_1 \mathcal{c}_2 < r_2+r_1$
$$
|3-| r|| < 5 < 3+|r|
$$
where, $C_1 C_2=\sqrt{(4-1)^2+(-1-3)^2}=5$
$$
\begin{aligned}
& =\left\{\begin{array}{c}
|3-| r \mid < 5 \\
3+|r|>5
\end{array}\right. \\
& =\left\{\begin{array}{c}
r \in(-8,8) \\
r \in(-\infty,-2) \cup(2, \infty)
\end{array}\right. \\
& =r \in(-8,-2) \cup(2,8)
\end{aligned}
$$