Show that the points (1,2) (3,-4) (5,-6) (19,8) are cancyclic and find the eqaution of the circle on which they lie

Equation of circle
$\begin{array}{l}{\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{C}=0\\ 1+4+2\mathrm{g}+4\mathrm{f}+\mathrm{C}=0\to \left(1\right)\\ 9+16+6\mathrm{g}+8\mathrm{f}+\mathrm{C}=0\to \left(2\right)\end{array}$
$25+36+10\mathrm{g}-12\mathrm{f}+\mathrm{C}=0\to \left(3\right)$
Subtracting (2) -(1) we get 20 + 4g -12f = 0
5 + g - 3f = 0→(4)
Soluing (4) and (5) we get
$\mathrm{f}=-2,\mathrm{g}=-11,\mathrm{c}=25$
equation of circle will be
${\mathrm{x}}^2+{\mathrm{y}}^2-22\mathrm{x}-4\mathrm{y}+25=0\to \left(6\right)$
Point (19,8) substituting in equation (4)
$(19{)}^2+(8{)}^2-22.19-4.8+25=0$
Hence (19,8) lie on circel and four points are con cyclic