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.

Another way to do this is to use the geometric fact that the perpendicular bisector of a chord passes through the center of a circle.  The line through P(5,7) and Q(6,6)I has slope (7- 6)/(5- 6)= -1 and midpoint (11/2, 13/2).  The perpendicular bisector is y= (x- 11/2)+ 13/2)= x+ 1.  The line through Q(6, 6) and R(2,−2) has slope (6-(-2))/(6- 2)= 8/4= 2 and midpoint (4, 2).  The perpendicular bisector is y= -(1/2)(x- 4)+ 2= -(1/2)x+ 4.  
Those two lines intersect when y= x+ 1= -(1/2)x+ 4 so (3/2)x= 3, x= 2.  Then y= 2+ 1= 3.  The center of this circle is (2, 3) and the radius is $\sqrt{(2- 6)^2+ (6- 3)^2}= \sqrt{25}= 5$.  The equation of that circle is $(x- 2)^2+ (y- 3)^2= 25$.  
As a check $(5- 2)^2+ (7- 3)^2= 9+ 16= 25$, $(6- 2)^2+ (6- 3)^2= 16+ 9= 25$, and $(2- 2)^2+ (-2- 3)^2= 0+ 25= 25$.
