Show that $\left(1, 1\right)$, $\left(-6, 0\right)$, $\left(-2, 2\right)$, $\left(-2, -8\right)$ are concyclic.
Let the given points be $A \left(1, 1\right)$, $B\left(-6, 0\right)$, $C \left(-2,2\right)$, $\left(-2, -8\right)$
By drawing a rough sketch it can be seen that for the quadrilateral $ABCD$,
$AB$ and $CD$ are the diagonals;
$AC, \; BD$ and $BC, \; DA$ are the opposite sides.
Now,
$AB = \sqrt{\left(1 + 6\right)^2 + 1^2} = \sqrt{50} = 5 \sqrt{2}$
$CD = \sqrt{\left(-2 + 2\right)^2 + \left(2 + 8\right)^2} = 10$
$\therefore \;$ Product of measure of diagonals $= AB \times CD = 5 \sqrt{2} \times 10 = 50 \sqrt{2}$ $\;\;\; \cdots \; (1)$
$AC = \sqrt{\left(1 + 2\right)^2 + \left(1 - 2\right)^2} = \sqrt{10}$
$BD = \sqrt{\left(-6 + 2\right)^2 + \left(0 + 8\right)^2} = \sqrt{80} = 4 \sqrt{5}$
$BC = \sqrt{\left(-6 + 2\right)^2 + \left(0 - 2\right)^2} = \sqrt{20} = 2 \sqrt{5}$
$DA = \sqrt{\left(1 + 2\right)^2 + \left(1 + 8\right)^2} = \sqrt{90} = 3 \sqrt{10}$
$\therefore \;$ Sum of the products of the measures of the pairs of opposite sides
$\begin{aligned}
= AC \times BD + BC \times DA & = \sqrt{10} \times 4 \sqrt{5} + 2 \sqrt{5} \times 3 \sqrt{10} \\\\
& = 20 \sqrt{2} + 30 \sqrt{2} \\\\
& = 50 \sqrt{2} \;\;\; \cdots \; (2)
\end{aligned}$
$\therefore \;$ We have from equations $(1)$ and $(2)$,
$AB \times CD = AC \times BD + BC \times DA$
i.e. $\;$ product of measures of diagonals $=$ sum of the products of the measures of the pairs of opposite sides
$\therefore \;$ By Ptolemy's theorem, a quadrilateral can be inscribed in a circle with the given points.
$\implies$ The given points are concyclic.