Show that the points (9,1) (7.9), (-2,12), (6,10) are concyclic and find equation of the circle on which they lie.

Equation of the circle is ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$
This circle passes through
$\mathrm{A}(9,1),\mathrm{B}(7,9),\mathrm{C}(-2,12),\mathrm{D}(6,10)$
$\begin{array}{l}81+1+18\mathrm{g}+2\mathrm{f}+\mathrm{c}=0\Rightarrow \\ 18\mathrm{g}+2\mathrm{f}+\mathrm{c}=-82\dots \dots .\left(1\right)\\ 49+81+14\mathrm{g}+18\mathrm{f}+\mathrm{c}=0\Rightarrow \\ 14\mathrm{g}+18\mathrm{f}+\mathrm{c}=-130\dots \dots \left(2\right)\\ 4+144-4\mathrm{g}+24\mathrm{g}+\mathrm{c}=0\Rightarrow \\ -4\mathrm{g}+24\mathrm{f}+\mathrm{c}=-148\dots \dots .\left(3\right)\end{array}$
$\text{ (1) - (2) gives }$
$\begin{array}{l}4\mathrm{g}-16\mathrm{f}=48\Rightarrow \mathrm{g}-4\mathrm{f}=12\dots \left(4\right)\\ \left(2\right)-\left(3\right)\text{ gives }\\ \begin{array}{rl}18\mathrm{g}-6\mathrm{f}=18\Rightarrow 3\mathrm{g}-\mathrm{f}=3& \Rightarrow \mathrm{f}=3\mathrm{g}-3\\ & \Rightarrow \mathrm{f}=3\mathrm{g}-3\dots .\left(5\right)\end{array}\end{array}$
sub.(5) in (4)
$\begin{array}{l}\mathrm{g}-4(3\mathrm{g}-3)=12\Rightarrow \mathrm{g}-12\mathrm{g}+12=12\\ \Rightarrow -11\mathrm{g}=0\Rightarrow \mathrm{g}=0\end{array}$
Substituting g = 0 in (5) f = -3
Substituting the values of g,f in (i)
$18\left(0\right)+2(-3)+\mathrm{c}+82=0\Rightarrow \mathrm{c}=6-82=-76$
Equation of the required circle is ${\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{y}-76=0$
subtitute D = (6,10) in the above eq.
$\begin{array}{l}{\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{y}-76=6^2+{10}^2-6\left(10\right)-76\\ =36+100-60-76\\ =136-136=0\end{array}$
D(6,10) lies on the circle passing thorugh A,B, C
$\therefore$A,B,C and D are concylic.
Equation of the circle is ${\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{y}-76=0$