finding equation of circle in complex plane So i was asked to find the equation of the circle going through 1, i, and 0
Now i know that the equation of circle in complex form is:
$|z - z_0| = r$ where $r$ is radius.
So based on these values, my idea was to obtain the radius and try and find the origin of my circle. Nice idea, but executing it did not come off. There was a solution provided, and i am trying to make the link between what the final solution is and how to get the origin of my circle and the radius. 
The solution was $|z - \frac{1+i}{2}| = \frac{1}{2^{1/2}}$
 A: You must solve the following system of equations:
$\left\{ \array{|c-0|=r\\|c-1|=r\\|c-i| =r}\right.$
where $c$ is the center of the circle and $r$ is the radius.
Assuming that $c=x+iy$, after evaluation we get the following system:
$\left\{ \array{(1-x)^2=x^2 \\ (1-y)^2 = y^2}\right.$
and then:
$\left\{ \array{1-2x=0 \\ 1-2y =0}\right.$
So $(x,y)= (\frac{1}{2}, \frac{1}{2})$ and $c=\frac{i+1}{2}$
Back to the radius we take any of the initial equations, for example the first one:
$|c-0|=r$
And solve it for $r$
$|c|=r$
$r=\frac{\sqrt{2}}{2}$
The equation of circle is then:
$|z-\frac{1+i}{2}| = \frac{\sqrt{2}}{2} $
A: The center of the circle must have form $z=x+ix$ for some $x\in\mathbb{R}$ since it must lies on the line which passes through origin and  perpendicular to line passing through $i$ and $1$.
And $r=|z-1|=|z|$ which gives you the value of $x$. Hence you find the center and you can get the radius $r$.
A: Let $c$ be the center of the circle and $r$ its radius, then $\,|c-0|=r\,$ so $\,c \bar c = r^2\,$ and also:
$$
\require{cancel}
\begin{cases}
\bcancel{r^2} = |c-1|^2=(c-1)(\bar c -1) = \bcancel{c \bar c} - (c+\bar c) +1 \\
\bcancel{r^2} = |c-i|^2 = (c-i)(\bar c + i) = \bcancel{c \bar c} + i(c - \bar c) +1
\end{cases}
$$
Therefore $c+\bar c=1$ and $c-\bar c = -1 / i = i$ thus $c=\frac{1}{2}\big((c+\bar c)+(c-\bar c)\big) = \cfrac{1+i}{2}\,$.
Then $\,r=|c|=\cfrac{\sqrt{2}}{2}\,$ so the equation of the circle is $\left|z-\cfrac{1+i}{2}\right|=\cfrac{\sqrt{2}}{2}\,$.
