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.
 A: 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 $g=-7/3$.  
A: Just a different approach: When you construct perpendicular bisectors of PQ and QR, then these lines intersect each other exactly in the center of the circle.
So midpoint of PQ is $(5.5,6)$, slope of line PQ is $-1$ and perpendicular bisector is then given by: $y-6=1(x-5.5)$. Midpoint of QR is $(4,2)$, slope of line QR is $2$ and perpendicular bisector is given by $y-2=2(x-4)$. Finding the intersection of these lines is easy by solving $x-5.5+6=2(x-4)+2$. Once you have the coordinates of this intersection, you have the center of the circle and the $r^2$ (radius squared) can be found by Pythagorean theorem which in essence gives you the equation of the circle. Can you solve it from here?
A: 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 $(2- 2)^2+ (-2- 3)^2= 0+ 25= 25$.
A: Another way could be successive elimination. Considering the three equations (please, notice that your third equation is wrong) $$10g+14f+c+74=0\tag 1$$ $$12g+12f+c+72=0\tag 2$$ $$4g-4f+c+8=0\tag 3$$ From $(1)$, eliminate $$c=-14 f-10 g-74\tag 4$$ Replace in $(2)$ and $(3)$ to get $$-2 f+2 g-2=0\tag 5$$ $$10 f+14 g+74=0 \tag 6$$ From $(5)$, eliminate $$f= g-1\tag 7$$ Replace in $(6)$ $$24 g+48=0\tag 8$$ So $g=-2$ and going backwards $f=-3$ and $c=-12$.
A: $\begin{vmatrix}
x^2+y^2&x&y&1\\
5^2+7^2&5&7&1\\
6^2+6^2&6&6&1\\
2^2+(-2)^2&2&-2&1
\end{vmatrix}=0$    
$-12(x^2+y^2)+48x+72y+144=0$
