Show that the tangent at (–1, 2) of the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-8\mathrm{y}+7=0$ touches the circle ${\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}+6\mathrm{y}=0$ and also find its point of contact.

Given circles are
$\begin{array}{l}{\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-8\mathrm{y}+7=0\\ {\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}+6\mathrm{y}=0\end{array}$
$\text{ Let }\mathrm{P}=\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=(-1,2)$
The equation of the tangent at P to (1) is
$\begin{array}{l}\mathrm{x}(-1)+\mathrm{y}\left(2\right)-2(\mathrm{x}-1)-4(\mathrm{y}+2)+7=0\\ \Rightarrow -3\mathrm{x}-2\mathrm{y}+1=0\Rightarrow 3\mathrm{x}+2\mathrm{y}-1=0 ...\left(3\right)\end{array}$
Centre of (2), C = (–2, –3)
radius of (2), $\mathrm{r}=\sqrt{4+9}=\sqrt{13}$
Perpendicular distance from C to (3)
$=\left|\frac{3(-2)+2(-3)-1}{\sqrt{13}}\right|=\frac{13}{\sqrt{13}}=\sqrt{13}=\mathrm{r}$
$\therefore$(3) touches (2)
Let Q(h, k) be the point of contact of (2) & (3)
$\therefore$ Q is the foot of the perpendicular from C to (3)
$\begin{array}{l}\therefore \frac{\mathrm{h}+2}{3}=\frac{\mathrm{k}+3}{2}=\frac{-(-13)}{13}\\ \Rightarrow \mathrm{h}=1,\mathrm{k}=-1\therefore \mathrm{Q}=(1,-1)\end{array}$