Finding the equation of a circle $A=(3,1)$ and $B=(-1,-1)$ are points on a circle of center $(k, -3k)$ find the value of $k$ 
I begin by assinging the values  $\ g = -k $ and  $\ f=3k $. 
I then substitute $(3, 1)$ and $ g= -k, f= 3k $ into $\ x^2+y^2+2gx+2fy+c$.  That leaves me with $c = 10$, meaning $k= -2$. Where have I gone wrong? I know its wrong.
FINAL EDIT:
Very sorry, as foolish as this sounds, there is a mix up in the numbering of the answers, I actually got the answer correct. Thank you for your time, sorry about wasting it. 
Please delete this Question.
 A: I'd approach the problem differently.  First, the perpendicular bisector of any chord of a circle contains the center of the circle, so I'd start by writing an equation for the perpendicular bisector of $\overline{AB}$.  Then, substitute $(k,-3k)$ into that equation and solve for $k$.
Using this method, I also get $k=-2$; how do you know that is wrong?
A: I don't know what that $K$ is and I don't care as I don't need to know it for my solution.
Issac has given an answer appealing to geometry. That you mention the equation of circle and seem to be working with it, this might be of some help as well:
So, the equation of the circle becomes, $$x^2+y^2-2kx+6ky+c=0$$
Now plugging in $(3,1)$, you'll have, 
$$9+1 -6k+6k+c=0 \implies c=-10$$
Note the negative sign here.
Now, plug in $(-1,-1)$, you'll have, $$1+1+2k-6k-10=0 \implies-4k-8=0 \implies k=-2$$
Hence, $\boxed{c=-10;k=-2}$.
Hope this helps.
A: Here's another variation. There will be a circle centered at $C=(k,3k)$ and going through $A$ and $B$ if and only if $A$ and $B$ are equidistant from $C$. Analitically, if and only if
$$
(3-k)^2+(1+3k)^2=(-1-k)^2+(-1+3k)^2.
$$
The equation simplifies to $8+4k=0$, i.e. $k=-2$.
A: Since ( k, -3k ) is the centre of the circle and the radius(r) of the circle is equal to r^2 = (k-3)^2 + (-3k-1)^2 and, r^2 = (k+1)^2 + (-3k+1)^2 ... by equating the two equations we have k=-2
