Show that x + y + 1 = 0 touches the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-3\mathrm{x}+7\mathrm{y}+14=0$ and find the point of contact

$\text{ Given circle is }{\mathrm{x}}^2+{\mathrm{y}}^2-3\mathrm{x}+7\mathrm{y}+14=0\dots \dots \left(1\right)$
$\text{ Centre }\mathrm{c}=\left(\frac{3}{2},\frac{-7}{2}\right)$

$\text{ Radius }\mathrm{r}=\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}=\sqrt{\frac{9}{4}+\frac{49}{4}-14}=\frac{1}{\sqrt{2}}$
$\text{ Given line is }\mathrm{x}+\mathrm{y}+1=0\text{...(2) }$
$\mathrm{D}={\perp }^{\mathrm{ar}}\text{ distance form centre }\mathrm{c}\left(\frac{3}{2},\frac{-7}{2}\right)\text{ to given }$$\text{ line }=\frac{\left|\frac{3}{2}-\frac{7}{2}+1\right|}{\sqrt{1^2+1^2}}=\frac{1}{\sqrt{2}}\Rightarrow \mathrm{d}=\mathrm{r}$
$\therefore$Given line touches the given circle for point of contact, solving (1) and (2)
y = -x -1 Put is (1)
${\mathrm{x}}^2+(-\mathrm{x}-1{)}^2-3\mathrm{x}+7(-\mathrm{x}-1)+14=0$
$\begin{array}{l}\Rightarrow {\mathrm{x}}^2+{\mathrm{x}}^2+2\mathrm{x}+1-3\mathrm{x}-7\mathrm{x}-7+14=0\\ \Rightarrow 2{\mathrm{x}}^2-8\mathrm{x}+8=0\mathrm{y}=-\mathrm{x}-1\end{array}$
${\mathrm{x}}^2-4\mathrm{x}+4=0\Rightarrow \mathrm{x}=2=-2-1=-3$
Point of contact P = (2, –3)