TITLE: How to find the center of the circle that contains three given complex numbers?
QUESTION [5 upvotes]: Suppose $\alpha_1, \alpha_2, \alpha_3 $ are complex numbers which are not collinear. Is it possible to use some geometry to find the center of the circle that contains $\alpha_1, \alpha_2, \alpha_3 $ ?

REPLY [2 votes]: Here is a solution with a somewhat different flavor. Using DFT we can write the three points $a_j$ in the form
$$a_j=m+p\omega^j+q\omega^{-j}\ ,$$
where $\omega:=e^{2\pi i/3}$, and
$$m:={1\over3}(a_1+a_2+a_3),\quad p:={1\over3}(a_1\bar\omega +a_2\omega+a_3),\quad
q:={1\over3}(a_1\omega +a_2\bar\omega+a_3)\ .\tag{1}$$
Now we have to choose the point $z$ in such a way that the quantity
$$|a_j-z|^2=(m+p\omega^j+q\omega^{-j}-z)(\bar m+\bar p\omega^{-j}+\bar q\omega^j-\bar z)\tag{2}$$
does not depend on $j$. This is the case iff the  coefficients  $c$ and $\bar c$ of $\omega^j$  and $\omega^{-j}$ resulting after computing the right hand side of $(2)$  are $=0$. Collecting the relevant terms we get
$$c=m\bar q+p\bar m-p\bar z+q\bar p-z\bar q\ .$$
This leads to the linear system
$$\eqalign{\bar q z+p\bar z&=\bar pq+p\bar m+\bar q m \cr
\bar p z+q\bar z&=p\bar q+\bar p m+ q \bar m \ . \cr}$$
Solving for $z$ and plugging in the values $(1)$ gives
$$z={(a_2-a_3)|a_1|^2+(a_3-a_1)|a_2|^2+(a_1-a_2)|a_3|^2 \over
(a_2-a_3)\bar a_1+(a_3-a_1)\bar a_2+(a_1-a_2)\bar a_3}\ .$$