Find circumcenter when distance between ABC points of triangle with two points's ratio given The complete problem is:
I am having three points A,B,C whose ratio of the  distances from points (1,0) and (-1,0) is 1:3 each. Then I need the coordinates of the circumcenter of the triangle formed by points A,B & C.
Can anybody tell me how to proceed?

I'd proceed like this:


*

*Prove that, in 2D, the set of all points defined by a fixed ratio
$\rho\neq1$ of their distances to two distinct given points
(which I will call poles for reasons not explained here)
is a circle $K$.
(I'd use equations with cartesian coordinates for that,
but if you find a coordinate-free reasoning, that would be interesting.)

*Conclude that the circumcenter of $\triangle ABC$ is the (yet unknown)
center $M$ of that circle $K$.

*Find two points $D,E$ on the line through the poles (here: the $x$-axis)
with the same given distance ratio $\rho$.

*Argue with symmetry and conclude that $M$ must be the midpoint of the
segment connecting $D$ and $E$. In particular, $M$ lies on the $x$-axis.


Update: An illustration:

More could be said on this ratio-and-circle topic, but
I have kept that to the comments section.
