Show that four common tangents can be drawn for the circles given by ${\mathrm{x}}^2+{\mathrm{y}}^2-14\mathrm{x}+6\mathrm{y}+33=0$, ${\mathrm{x}}^2+{\mathrm{y}}^2+30\mathrm{x}-2\mathrm{y}+1=0$ and find the internal and external centres of similitudes 

Given circles are
$\begin{array}{cc}{\mathrm{x}}^2+{\mathrm{y}}^2-14\mathrm{x}+6\mathrm{y}+33=0,& {\mathrm{x}}^2+{\mathrm{y}}^2+30\mathrm{x}-2\mathrm{y}+1=0\\ \mathrm{Centre}{\mathrm{C}}_1=(-\mathrm{g},-\mathrm{f})& \mathrm{Centre} {\mathrm{C}}_2=(-\mathrm{g},-\mathrm{f})\\ =(7,-3)& =(-15,1)\\ \begin{array}{l} \mathrm{Radius} {\mathrm{r}}_2=\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}\\ \begin{array}{c}=\sqrt{49+9-33}\\ =\sqrt{25}=5\end{array}\end{array}& \begin{array}{l} \mathrm{Radius} {\mathrm{r}}_2=\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}\\ \begin{array}{c}=\sqrt{225+1-1}\\ =15\end{array}\end{array}\end{array}$
Distance between the centres
$\begin{array}{l}{\mathrm{C}}_1{\mathrm{C}}_2=\sqrt{484+16}=\sqrt{500}\\ {\mathrm{r}}_1+{\mathrm{r}}_2=5+15=20=\sqrt{400}\\ \sqrt{500}>\sqrt{400}\Rightarrow {\mathrm{C}}_1{\mathrm{C}}_2>{\mathrm{r}}_1+{\mathrm{r}}_2\end{array}$

$\therefore$The circles disjoint
$\therefore$ 2 Direct common tangents and
2 transverse common tangents exist
$\therefore$ No. of common tangents n = 4
Internal center of similituds (ICS) divides C₁C₂ in the ratio r₁ : r₂ internally
$\begin{array}{l}=5:15=1:3\\ \therefore \mathrm{ICS}=\frac{1\cdot {\mathrm{C}}_2+3{\mathrm{C}}_1}{1+3}=\frac{1(-15,1)+3(7,-3)}{4}=\frac{(-15,1)+(21,-9)}{4}=\left(\frac{-15+21}{4},\frac{1-9}{4}\right)=\left(\frac{6}{4},\frac{-8}{4}\right)=\left(\frac{3}{2},-2\right)\end{array}$