Find the circumcentre of a triangle with points $z_1,z_2$ and the origin as its vertices where $z_1,z_2 \in \mathbb{C}$ Let $z_1,z_2$ be two non-zero complex numbers satisfying $$\lvert z_1+z_2 \rvert=\lvert z_1-z_2 \rvert$$

I have done many things while trying to find this one out.
$\lvert z_1+z_2 \rvert^2= \lvert z_1-z_2 \rvert^2$
$\implies \lvert z_1 \rvert^2 +\lvert z_2 \rvert^2 + 2\lvert z_1 \rvert\cdot \lvert z_2 \rvert\cdot cos(\theta_1-\theta_2)=\lvert z_1 \rvert^2 +\lvert z_2 \rvert^2 - 2\lvert z_1 \rvert\cdot \lvert z_2 \rvert\cdot cos(\theta_1-\theta_2)$
$\implies cos(\theta_1-\theta_2)=0$
$\implies\theta_1-\theta_2 =\dfrac{\pi}{2} $
The three sides of the triangle are :
$ \lvert z_1 \rvert $,$ \lvert z_2 \rvert $,$ \lvert z_1-z_2 \rvert$ respectively.
And I did several other things like I assumed $z_1=x_1+iy_1$ & $z_2=x_2+iy_2$
And by using the relation $\lvert z_1+z_2 \rvert=\lvert z_1-z_2 \rvert$, we get
$x_1x_2=y_1y_2$ 
I then considered the vertices of the triangle be like $(0,0),(1,2),(2,1)$ where $x_1=1,x_2=2 \text {&} y_1=2,y_2=1$ But this does not give me the answer $\dfrac{ (z_1+z_2)}{2}$
Please give me a generic way to solve this.
 A: As I showed this earlier and @dxiv has enlightened my insight,
$\lvert z_1+z_2 \rvert^2= \lvert z_1-z_2 \rvert^2$
$\implies \lvert z_1 \rvert^2 +\lvert z_2 \rvert^2 + 2\lvert z_1 \rvert\cdot \lvert z_2 \rvert\cdot cos(\theta_1-\theta_2)=\lvert z_1 \rvert^2 +\lvert z_2 \rvert^2 - 2\lvert z_1 \rvert\cdot \lvert z_2 \rvert\cdot cos(\theta_1-\theta_2)$
$\implies cos(\theta_1-\theta_2)=0$
$\implies\theta_1-\theta_2 =\dfrac{\pi}{2} $
The triangle is right angled at the origin as $ z_1 $,$z_2$ sides make an angle of $\dfrac{\pi}{2}$ with each other at the origin.
Thus the hypotenuse is  $ z_1 - z_2$ and in a right angled triangle, the circumcenter is at the midpoint of the hypotenuse
Thus the circumcentre is $\dfrac{1}{2}.(z_1+z_2)$
And as @dxiv has pointed To complete the proof, take  $ z=\dfrac{1}{2}.(z_1+z_2)$ and prove that $\lvert z \rvert= \dfrac{1}{2}.\lvert z_1-z_2 \rvert$
Proved.
A: Let $z_1=(x_1,y_1)$ and  $z_2=(x_2,y_2)$ ans $0 = (0,0,0)$
The circumcenter $C=(c_1,c_2)$ satisfies $$ |C-0|=|C-z_1|= |C-z_2|$$
That us $$ c_1^2+c_2^2 =(c_1-x_1)^2 + (c_2-y_1)^2 =(c_1-x_2)^2 + (c_2-y_2)^2$$
Upon some cancellations we have a system to solve for $c_1$ and $c_2$
$$2c_1x_1+2c_2y_1=x_1^2+y_1^2\\2c_1x_2+2c_2y_2=x_2^2+y_2^2$$
A: Since $$|z_1- z_2| = |z_1+z_2|$$ we see that $z_1$ is on perpendicular bisector of segment between $z_1$ and $z_2$, so angle $\angle z_10z_2 = 90^{\circ}$ which means that circumcenter is a midpoint of segment between $z_1$ and $z_2$. So $$\zeta ={z_1+z_2\over 2}$$  where $\zeta $ is circumcenter. 
