Question
Show that the points $(3, -2), (1, 0), (-1, -2)$ and $(1, -4)$ are concyclic.

Answer

we have,
$P = (3, -2), Q = (1,0), R = (-1,-2)$ and $S = (1, -4)$
let us consider A circle $x^2 + y^2 + 2gx + 2fy + c = 0 ........ (1)$
Passes through $P, Q$ & $R$
$\therefore 9 + 4 + 6g - 4f + c = 0 ........ (2)$
$1 + 0 + 2g - 0 + c = 0 ............ (3)$
$1 + 4 - 2g - 4f + c = 0 ............ (4)$
Solving $(2), (3)$ & $(4) $we get,
$g = -1, f = 2$ & $c = 1$
from(1)
The required equation of ercle is
$x^2 + y^2 - 2x + 4y + 1 = 0 ......... (5)$
Clearly $s = (1, -4)$ satisfy $(5)$
Thus,
$P, Q, R$ & $S$ are concydic

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