Equation for orthocentre in argand plane

Question

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.

Answer 1

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.