Intersection of three circles each with the same unknown radius I'm kind of struggling with this one;

You are given points $A(100, 42)$, $B(33,74)$ and $C(-26,6)$. Point $D$ has unknown coordinates but the distance from $D$ to $A$ is the same as that from $D$ to $B$ and that from $D$ to $C$. Determine the coordinates of $C$.

At first, I thought it would just be calculating the intersections of the three circles you can create, all three with $r = d(A,C)$, but it didn't work out too well.
Can anybody please help?
 A: Assuming that you mean that is $D$ instead of $C$ the point whose coordinates are to be determined, note that $D$ is the circumcenter of the triangle $ABC$.
There are two basic ways to find it.
(1):
Find the equations of two of the bisectors of the sides of the triangle, and find their intersection. This is $D$.
(2):
Consider a circle $x^2+y^2+ax+by+c=0$. Substitute the coordinates of $A$, $B$ and $C$ in this equation to obtain a $3\times3$ linear system on the unknowns $a$, $b$ and $c$. Solve it. $D$ is the center of this circle.
Remark: The coordinates of $A$, $B$ and $C$ seem big. If you don't have to make the calculations by hand, you should use software. Geogebra is great for this.
A: Personally, I will think with vectors. It will end up at the same place nonetheless. Assuming you are finding coordinates of $D$.
We have $$\left|\overrightarrow{AD}\right|=\left|\overrightarrow{BD}\right|=\left|\overrightarrow{CD}\right|$$
with $\overrightarrow{OA}=\binom{100}{42}$, $\overrightarrow{OB}=\binom{33}{74}$, and $\overrightarrow{OC}=\binom{-26}{6}$.
Let $\overrightarrow{OC}=\binom{x}{y}$.
We have 
$$(x-100)^2+(y-42)^2=(x-33)^2+(y-74)^2=(x+26)^2+(y-6)^2$$
This seems very complicated but it will be simpler if we do two at a time.
Simplifying pairwise, we have
$$y= \frac{67}{32}x-\frac{5199}{64}$$
$$y= -\frac{59}{68}x+\frac{5853}{136}$$
$$y= -\frac{7}{2}x-\frac{307}{2}$$
In fact, solving any two of the linear equations will give the result if the question is valid.
So we have
$$D\equiv\left(\frac{15023}{358},\frac{4745}{716}\right)$$
which, to be honest, is an unnecessarily ugly answer.
Diagram:

