Find the transverse common tangents of the circles
${\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-10\mathrm{y}+28=0 \mathrm{and} {\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}-6\mathrm{y}+4=0$

Given equation of the circles are
$\begin{array}{l}{\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-10\mathrm{y}+28=0\to \left(1\right)\\ {\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}-6\mathrm{y}+4=0\to \left(2\right)\end{array}$
For the circle (1) centre C₁ = (2,5)
$\text{ Radius }{\mathrm{r}}_1=\sqrt{(-2{)}^2+(-5{)}^2-28}$
${\mathrm{r}}_1=\sqrt{4+25-28}{\mathrm{r}}_1=1$
$\text{ For the circles (2), centre, }{\mathrm{C}}_2=(-2,3)$
$\begin{array}{l} \mathrm{Radius}\cdot {\mathrm{r}}_2=\sqrt{(2{)}^2+(-3{)}^2-4}\\ =\sqrt{4+9-4}\end{array}$
${\mathrm{r}}_2=3$
$\begin{array}{l}{\mathrm{C}}_1{\mathrm{C}}_2=\sqrt{(-2-2{)}^2+(5-3{)}^2}\\ =\sqrt{(-4{)}^2+(2{)}^2}\\ {\mathrm{C}}_1{\mathrm{C}}_2=\sqrt{20}\end{array}$
$\text{ Now }{\mathrm{r}}_1+{\mathrm{r}}_2=1+3=4=\sqrt{16}$
$\because {\mathrm{C}}_1{\mathrm{C}}_2>{\mathrm{r}}_1+{\mathrm{r}}_2$
Given circle are each circle lies completely out side the other
Let A₁ be the internal centre of similitude .
The internal centre of similitude A₁ divides
C₁C₂ in the ratio r₁ : r₂ (1: 3) internally
$\because$ The internal center of similitude
$\begin{array}{l}{\mathrm{A}}_1=\left(\frac{{\mathrm{mx}}_2+{\mathrm{mx}}_1}{\mathrm{m}+\mathrm{n}},\frac{{\mathrm{my}}_2+{\mathrm{my}}_1}{\mathrm{m}+\mathrm{n}}\right)\\ =\left(\frac{1(-2)+3\left(2\right)}{1+3},\frac{1(3+3(5\left)\right)}{1+3}\right)\\ =\left(\frac{-2+6}{4},\frac{3+15}{4}\right)=\left(\frac{4}{4},\frac{18}{4}\right)\\ =\left(1,\frac{9}{2}\right)\end{array}$
The equation to the pair of transverse common tangents is $\mathrm{S}\cdot {\mathrm{S}}_{11}={\mathrm{S}}_1^2$
$\begin{array}{l}\left({\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}\right)\\ \left({\mathrm{x}}_1^2+{\mathrm{y}}_1^2+2{\mathrm{gx}}_1+2{\mathrm{fy}}_1+\mathrm{c}\right)\\ ={\left({\mathrm{xx}}_1+{\mathrm{yy}}_1+\mathrm{g}\left(\mathrm{x}+{\mathrm{x}}_1\right)+\mathrm{f}\left(\mathrm{y}+{\mathrm{y}}_1\right)+\mathrm{c}\right)}^2\\ \Rightarrow \left({\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-10\mathrm{y}+28\right)\\ \left(1^2+{\left(\frac{9}{2}\right)}^2+2(-2)\left(1\right)+2(-5)\left(\frac{9}{2}\right)+28\right)\\ ={\left[\mathrm{x}\left(1\right)+\mathrm{y}\left(\frac{9}{2}\right)-2(\mathrm{x}+1)-1-5\left(\mathrm{y}+\frac{9}{2}\right)+28\right]}^2\\ \Rightarrow \left({\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-10\mathrm{y}+28\right)\\ \left(1+\frac{81}{4}-4-45+28\right)\end{array}$
$\begin{array}{l}={\left(\mathrm{x}+\frac{9\mathrm{y}}{2}-2\mathrm{x}-2-5\mathrm{y}-\frac{5}{2}+28\right)}^2\\ \Rightarrow \left({\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-10\mathrm{y}+28\right)\end{array}$
$\begin{array}{l}\left(\frac{81}{4}-20\right)={\left(-\mathrm{x}-\frac{\mathrm{y}}{2}+\frac{7}{2}\right)}^2\\ \Rightarrow \left({\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-10\mathrm{y}+28\right)\left(\frac{1}{4}\right)\\ ={\left(\frac{-2\mathrm{x}-\mathrm{y}+7}{2}\right)}^2\\ \Rightarrow \left({\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-10\mathrm{y}+28\right)\\ =(-2\mathrm{x}-\mathrm{y}+7{)}^2\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-10\mathrm{y}+28\\ =4{\mathrm{x}}^2+{\mathrm{y}}^2+49+4\mathrm{xy}-14\mathrm{y}-28\mathrm{x}\\ \Rightarrow 3{\mathrm{x}}^2+4\mathrm{xy}-24\mathrm{x}-4\mathrm{y}+21=0\end{array}$
Consider
$\begin{array}{l}3{\mathrm{x}}^2+4\mathrm{xy}=0\Rightarrow \mathrm{x}(3\mathrm{x}+4\mathrm{y})=0\\ \mathrm{x}=0\text{ and }3\mathrm{x}+4\mathrm{y}=0\\ \therefore 3{\mathrm{x}}^2+4\mathrm{xy}-24\mathrm{x}-4\mathrm{y}+21=(\mathrm{x}+\mathrm{\ell })\\ (3\mathrm{x}+4\mathrm{y}+\mathrm{k})\end{array}$
Comparing coefficient of x on both sides
$\text{ we get }\mathrm{k}+3\mathrm{\ell }=-24\to \left(3\right)$
Camparing coefficient of ‘x’ on both sides
$\text{ we get }4\mathrm{\ell }=-4\mathrm{\ell }=-1$
Substitude the volue of $\mathrm{\ell }$ in (3)
$\begin{array}{l}\Rightarrow \mathrm{k}+3(-1)=-24\\ \Rightarrow \mathrm{k}-3=-24\\ \Rightarrow \mathrm{k}=-24+3\\ \to \mathrm{k}=-21\end{array}$
$\therefore \text{ The equation of the transverse common tangents are }$
$\mathrm{x}-1=0\text{ and }3\mathrm{x}+4\mathrm{y}-21=0$