Given circles $x^2+y^2+6 x+8 y+24=0$ and $x^2+y^2$
$$
\begin{aligned}
& -6 x-8 y+9=0 \\
& \Rightarrow x^2+y^2+6 x+8 y+9+24-9+16-16=0 \\
& (x+3)^2+(y+4)^2+24-9-16=0 \\
& (x+3)^2+(y+4)^2=(1)^2 \\
& \mathrm{r}_1=1, \text { centre } O_1=(-3,-4) \\
& \Rightarrow x^2+y^2-6 x-8 y+9=0 \\
& x^2+y^2-6 x-8 y+9+16-16=0 \\
& (x-3)^2+(y-4)^2=(16)=(4)^2 \\
& r_2=4 \text {, centre } O_2=(3,4)
\end{aligned}
$$
Here, $\mathrm{C}_1$ point divides $\mathrm{O}_1$ and $\mathrm{O}_2$ externally in their radius ratio. So, $r_1: r_2=1: 4$
$(3,4)$
$$
\begin{aligned}
& x_1=\frac{1 \times 3+4 \times-3}{1+4}=\frac{3-12}{5}=\frac{-9}{5} \\
& y_1=\frac{1 \times 4+4 \times-4}{1+4}=\frac{4-16}{5}=\frac{-12}{5} \\
& C_1\left(x_1, y_1\right) \rightarrow\left(\frac{-9}{5}, \frac{-12}{5}\right)
\end{aligned}
$$
Now,
$(3,4)$
$$
\begin{aligned}
& x_2=\frac{1 \times 3-4 \times(-3)}{1-4}=\frac{3+12}{-3}=\frac{15}{-3}=-5 \\
& y_2=\frac{1 \times 4-4 \times(-4)}{1-4}=\frac{4+16}{-3}=\frac{-20}{3} \\
& \mathrm{C}_2\left(x_2, y_2\right) \rightarrow\left(-5, \frac{-20}{5}\right)
\end{aligned}
$$
Distance of $\mathrm{C}_1 \mathrm{C}_2$
$$
\begin{aligned}
& =\sqrt{\left(\frac{-9}{5}+5\right)^2+\left(\frac{-20}{3}+\frac{12}{5}\right)^2}=\sqrt{\left(\frac{16}{5}\right)^2+16\left(\frac{16}{15}\right)^2} \\
& =\sqrt{\left(\frac{16}{5}\right)^2\left(1+\frac{16}{9}\right)}=\sqrt{\left(\frac{16}{5}\right)^2 \times \frac{25}{9}}=\frac{16}{5} \times \frac{5}{3}=\frac{16}{3}
\end{aligned}
$$