Equation for orthocentre in argand plane I would like to know the general equation for orthocentre (and other centres of the triangle as well, excluding the centroid, if possible) in the argand plane. In my case, I faced some difficulty in a particular problem where all vertices of the triangle lie on a circle. So, please provide me with the solving methodology, if a separate one exists for this case.
 A: Lemma. Let $O$ be the circumcentre of $\triangle ABC$. Then $\overrightarrow{OH}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}$.
Proof. You can do this yourself -- just verify that $\overrightarrow{AH}\cdot \overrightarrow{BC}=0$ etc.
So in complex numbers, if your vertices are $a$, $b$, $c$, circumcentre is $o$ and orthocentre $h$, then
$$(h-o)=(a-o)+(b-o)+(c-o)\implies h=a+b+c-2o.$$
In general, the circumcentre of $a$, $b$, $c$ is given by the fearsome determinant formula:
$$\begin{vmatrix}
a&a\overline{a}&1\\
b&b\overline{b}&1\\
c&c\overline{c}&1
\end{vmatrix}\div
\begin{vmatrix}
a&\overline{a}&1\\
b&\overline{b}&1\\
c&\overline{c}&1
\end{vmatrix}.$$
The proof of this formula is to consider the system
\begin{align*}
\lvert a-o\rvert^2&=R^2\\
\lvert b-o\rvert^2&=R^2\\
\lvert c-o\rvert^2&=R^2.
\end{align*}
Expand the moduli in terms of conjugates, then apply Cramer's rule where the unknowns are $o$, $\overline{o}$ and $R^2-o\overline{o}$.
This circumcentre formula becomes much more tractable when $c=0$ -- we just get $$\frac{ab\overline{(a-b)}}{\overline{a}b-a\overline{b}}.$$
So if you want to compute an arbitrary circumcentre, shift one of the vertices to the origin, apply the above (simpler) formula, then shift back.
