Find the equation of the direct common tangents to the circles
$\begin{array}{l}{\mathrm{x}}^2+{\mathrm{y}}^2+22\mathrm{x}-4\mathrm{y}-100=0\text{ and }\\ {\mathrm{x}}^2+{\mathrm{y}}^2-22\mathrm{x}+4\mathrm{y}+100=0\end{array}$

Given circles are ${\mathrm{x}}^2+{\mathrm{y}}^2+22\mathrm{x}-4\mathrm{y}-100=0,$ ${\mathrm{x}}^2+{\mathrm{y}}^2-22\mathrm{x}+4\mathrm{y}+100=0$
$\text{Centre }{\mathrm{C}}_1=(-11,2)$
$\text{Centre }{\mathrm{C}}_2={\mathrm{C}}_2=(-\mathrm{g},\mathrm{f})=(11,-2)$
$\text{Radius }{\mathrm{r}}_1=\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}$
$\text{Radius }{\mathrm{r}}_2=\sqrt{121+4-100}$
$\begin{array}{l}=\sqrt{121+4+100}=\sqrt{25}=5\\ =\sqrt{225}=15\end{array}$

Distance between the centres
$\begin{array}{l}{\mathrm{C}}_1{\mathrm{C}}_2=\sqrt{(22{)}^2+4^2}=\sqrt{500}\\ {\mathrm{r}}_1+{\mathrm{r}}_2=15+5=20=\sqrt{400}\\ \sqrt{500}>\sqrt{400}\Rightarrow {\mathrm{C}}_1{\mathrm{C}}_2>{\mathrm{r}}_1+{\mathrm{r}}_2\end{array}$

$\therefore$The circles are disjoint 2 Direct common tangents exist
External center of similitude divides C₁C₂ in the ratio r₁ : r₂
$\text{externally }=-{\mathrm{r}}_1:{\mathrm{r}}_2=-15:5=-3:1$
$\therefore \mathrm{ECS}=\frac{-3\cdot {\mathrm{C}}_2+1\cdot {\mathrm{C}}_1}{-3+1}$
$=\left(\frac{-33-11}{-2},\frac{6+2}{-2}\right)=(22,-4)$
Direct common tangents pass through ECS, let slope of DCT is m Equation of DCT is
$\begin{array}{l}\mathrm{y}-{\mathrm{y}}_1=\mathrm{m}\left(\mathrm{x}-{\mathrm{x}}_1\right)\\ \Rightarrow \mathrm{y}+4=\mathrm{m}(\mathrm{x}-22)\Rightarrow \mathrm{mx}-\mathrm{y}-22\mathrm{m}-4=0\end{array}$
$\text{But, it is a tangent to circle (1) }\Rightarrow \mathrm{d}={\mathrm{r}}_1\text{ i.e., }$
$\text{Length of }{\perp }^{\text{er }}\text{ from }{\mathrm{C}}_1(-11,2)\text{ to }\mathrm{DCT}={\mathrm{r}}_1$
$\begin{array}{l}\Rightarrow \frac{\left|\mathrm{m}\right(-11)-2-22\mathrm{m}-4|}{\sqrt{{\mathrm{m}}^2+1}}=15\\ \Rightarrow |-33\mathrm{m}-6|=15\sqrt{{\mathrm{m}}^2+1}\\ \Rightarrow 3|11\mathrm{m}+2|=15\sqrt{{\mathrm{m}}^2+1}\end{array}$
SOBS
$\begin{array}{l}(11\mathrm{m}+2{)}^2=25\left({\mathrm{m}}^2+1\right)\\ \Rightarrow 121{\mathrm{m}}^2+44\mathrm{m}+4=25{\mathrm{m}}^2+25\end{array}$
$\begin{array}{l}\Rightarrow 96{\mathrm{m}}^2+44\mathrm{m}-21=0\\ \Rightarrow 96{\mathrm{m}}^2+72\mathrm{m}-28\mathrm{m}-21=0\\ \Rightarrow 24\mathrm{m}(4\mathrm{m}+3)-7(4\mathrm{m}+3)=0\\ \Rightarrow (4\mathrm{m}+3)(24\mathrm{m}-7)=0\Rightarrow \mathrm{m}\frac{-3}{4},\mathrm{m}=\frac{7}{24}\end{array}$
$\text{If }\mathrm{m}=\frac{-3}{4}$
$\text{Equation of DCT is }\mathrm{y}+4=\frac{-3}{4}(\mathrm{x}-22)$
$\Rightarrow 4\mathrm{y}+16=-3\mathrm{x}+66\Rightarrow 3\mathrm{x}+4\mathrm{y}-50=0$
$\text{If }\mathrm{m}=\frac{7}{24}$
$\text{Equation of DCT is }\mathrm{y}+4=\frac{7}{24}(\mathrm{x}-22)$
$\Rightarrow 24\mathrm{y}+96=7\mathrm{x}-154\Rightarrow 7\mathrm{x}-24\mathrm{y}-250=0$