Find the equation of the circle passing ... Find the equation of the circle passing through the points $P(5,7)$, $Q(6,6)$ and $R(2,-2)$.
My Attempt:
Let the equation of the circle be:
$$x^2+y^2+2gx+2fy+c=0$$
The point $P(5,7)$ lies on the circle then,
$$5^2+7^2+10g+14f+c=0$$
$$10g+14f+c=-74$$-----(1)
The point $Q(6,6)$ lies on the circle then,
$$6^2+6^2+12g+12f+c=0$$
$$12g+12f+c=-72$$-------(2)
The point $R(2,-2)$ lies on the circle the,
$$2^2+(-2)^2+4g-4f+c=0$$
$$4g-4f+c=0$$-----(3).
Now, please help me from here.

Subtract $2^{nd}$ equation from $3$ times the $3^{rd}$ to get 
$$
  c-12f=36
$$  
Then subtract $2$ times the first from $5$ times the third to get:
$$
  3c-48f=148
$$ 
From the last two, subtract $3$ times the first from the second to get $f=-10/3$. Plugging back in one them gives $c=-4$. Now pluging in one of the first three, gives $g=-7/3$.  
