Show that the four points (–6, 0), (–2, 2), (–2, –8) and (1, 1) are concylic.

Given points are
$\mathrm{A}(-6,0),\mathrm{B}(-2,2),\mathrm{C}(1,1)\mathrm{\&}\mathrm{D}(-2,-8)$
Let equation of the circle passing through A,B,C
be ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$
A(–6,0) lies on it
$\begin{array}{l}\Rightarrow 36+0+2\mathrm{g}(-6)+2\mathrm{f}\left(0\right)+\mathrm{c}=0\\ \Rightarrow -12\mathrm{g}+\mathrm{c}=-36\dots .\left(1\right)\end{array}$
B(–2,2) lies on it
$\begin{array}{l}\Rightarrow 4+4+2\mathrm{g}(-2)+2\mathrm{f}\left(2\right)+\mathrm{c}=0\\ \Rightarrow -4\mathrm{g}+4\mathrm{f}+\mathrm{c}=-8\dots \dots \left(2\right)\end{array}$
C (1,1) lies on it
$\begin{array}{l}\Rightarrow 1^2+1^2+2\mathrm{g}\left(1\right)+2\mathrm{f}\left(1\right)+\mathrm{c}=0\\ \Rightarrow 2\mathrm{g}+2\mathrm{f}+\mathrm{c}=-2\dots .\left(3\right)\end{array}$
Eliminating ‘f’ from (2) & (3),
$\begin{array}{l}1\times \left(2\right)\Rightarrow -4\mathrm{g}+4\mathrm{f}+\mathrm{c}=-8\\ 2\times \left(3\right)\Rightarrow 4\mathrm{g}+4\mathrm{f}+2\mathrm{c}=-4\\ (-)-8\mathrm{g}+0-\mathrm{c}=-4\\ \Rightarrow -8\mathrm{g}-\mathrm{c}=-4\dots \dots ..\left(4\right)\end{array}$
Solving (4) & (1)
$-8\mathrm{g}-\mathrm{c}=-4\mathrm{\&}-12\mathrm{g}+\mathrm{c}=-36$
Adding $-20\mathrm{g}=-40\Rightarrow \mathrm{g}=2$
Put g,c in (3) : 
$2\left(2\right)+2\mathrm{f}-12=-2\Rightarrow 2\mathrm{f}=-2+8=6\Rightarrow \mathrm{f}=3$

$\therefore$Equation of the circle passing through A,B,C is
${\mathrm{x}}^2+{\mathrm{y}}^2+2\left(2\right)\mathrm{x}+2\left(3\right)\mathrm{y}-12=0$
$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}+6\mathrm{y}-12=0$
If D lies on the above circle, A,B,C,D are concyclic
LHS =$(-2{)}^2+(-8{)}^2+4(-2)+6(-8)-12$
$=4+64-8-48-12=68-68=0=\mathrm{LHS}$
D( -2, -8)  lies on the circle
$\therefore \mathrm{A},\mathrm{B},\mathrm{C},\mathrm{D} \mathrm{are} \mathrm{concyclic}$