In first circle,

${\left(x-2\right)}^2+{\left(y-3\right)}^2=25$

${\left(x-2\right)}^2+{\left(y-3\right)}^2=5^2 \dots \left(1\right)$

Compare equation $\left(1\right)$ with,

${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

$h=2, k=3 \text{\&} r_1=5$

Centre of circle $\left(1\right)$ is $\left(h,k\right)=\left(2,3\right)$

And it's radius is $=5 \mathrm{unit}$

For Second circle,

$25x^2+25y^2-40x-70y-160=0$

$25\left(x^2+y^2-\frac{8}{5}x-\frac{14}{5}y-\frac{32}{5}\right)=0$

$x^2+y^2-\frac{8}{5}x-\frac{14}{5}y-\frac{32}{5}=0$

$\Rightarrow {\left(x-\frac{4}{5}\right)}^2+{\left(y-\frac{7}{5}\right)}^2=\frac{32}{5}+\frac{16}{25}+\frac{49}{25}$

$\Rightarrow {\left(x-\frac{4}{5}\right)}^2+{\left(y-\frac{7}{5}\right)}^2=3 \dots \left(2\right)$

Comparing it with equation ${\left(x-h^{\text{'}}\right)}^2+{\left(y-k^{\text{'}}\right)}^2={\left(r^{\text{'}}\right)}^2$

$h^{\text{'}}=\frac{4}{5}, k^{\text{'}}=\frac{7}{5} \text{\&} r_2=3$

Centre of circle is $\left(h^{\text{'}},k^{\text{'}}\right)=\left(\frac{4}{5},\frac{7}{5}\right)$

Its Radius is $r_2=3 \mathrm{unit}$.

Point of Contact is $\left(\alpha ,\beta \right)=\left(\frac{r_1{x}_2-r_2{x}_1}{r_1-r_2},\frac{r_1{y}_2-r_2{y}_1}{r_1-r_2}\right)$

$\Rightarrow \left(\alpha ,\beta \right)=\left(\frac{5\times {\displaystyle \frac{4}{5}}-3\times 2}{5-3},\frac{5\times {\displaystyle \frac{7}{5}}-3\times 3}{5-3}\right)=\left(-1,-1\right)$

$\therefore \alpha +\beta =-1-1=-2$