If (1,2) (3,-4) (5,-6) and (c,8) are concyclic Then find c =

Let the equation of the circle
${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0\to \left(1\right)$
Since (1) is passing through (1,2)
$\begin{array}{l}1+4+2\mathrm{g}+4\mathrm{f}+\mathrm{c}=0\\ 2\mathrm{g}+4\mathrm{f}+\mathrm{c}=-5\to \left(2\right)\end{array}$
Since (1) is passing through (3,-4)
$\begin{array}{l}9+16+6\mathrm{g}-8\mathrm{f}+\mathrm{c}=0\\ 6\mathrm{g}-8\mathrm{f}+\mathrm{c}=-25\to \left(3\right)\end{array}$
Since (1) is passing through (5,-6)
$\begin{array}{l}25+36+10\mathrm{g}-12\mathrm{f}+\mathrm{c}=0\\ 10\mathrm{g}-12\mathrm{f}+\mathrm{c}=-61\to \left(4\right)\end{array}$
Solving (2) and (3)
$\begin{array}{l}2\mathrm{g}+4\mathrm{f}+\mathrm{c}=-5\\ 6\mathrm{g}-8\mathrm{f}+\mathrm{c}=-25\\ -4\mathrm{g}+12\mathrm{f}=20;\mathrm{g}-3\mathrm{f}+5=0\to \left(5\right)\end{array}$
Solving (3) and (4)
$\begin{array}{l}6\mathrm{g}-8\mathrm{f}+\mathrm{c}=-25\\ 10\mathrm{g}-12\mathrm{f}+\mathrm{c}=-61\\ -4\mathrm{g}+4\mathrm{f}=36\\ -\mathrm{g}+\mathrm{f}=9;\mathrm{g}-\mathrm{f}=-9\\ \mathrm{g}-\mathrm{f}+9=0\to \left(6\right)\end{array}$
Solving (5) and (6)
$\begin{array}{r}\begin{array}{lccc} & \mathrm{g}& \mathrm{f}& 1\\ -3& 5& 1& -3\end{array}\\ \begin{array}{llll}-1& 9& 1& -1\end{array}\\ \frac{\mathrm{g}}{-27+5}=\frac{\mathrm{f}}{5-9}=\frac{1}{-1+3}\end{array}$
$\frac{\mathrm{g}}{-22}=\frac{\mathrm{f}}{-4}=\frac{1}{2}\mathrm{g}=-11,\mathrm{f}=-2$
From (2)
$\begin{array}{l}-22-8+\mathrm{c}=-5\\ -30+\mathrm{c}=-5\\ \mathrm{c}=25\end{array}$
$\because \text{ Equation of the circle }{\mathrm{x}}^2+{\mathrm{y}}^2-22\mathrm{x}-4\mathrm{y}+25=0$
Given four points are con cyclic
(C,8) lies on the circle
$\begin{array}{l}{\mathrm{c}}^2+64-22\mathrm{c}-32+25=0\\ {\mathrm{c}}^2-22\mathrm{c}+57=0\\ {\mathrm{c}}^2-19\mathrm{c}-3\mathrm{c}+57=0\\ \mathrm{c}(\mathrm{c}-19)-3(\mathrm{c}-19)=0\\ (\mathrm{c}-3)(\mathrm{c}-19)=0\\ \mathrm{c}=19,3\end{array}$