The circumscribed centers are on a line Let $\mathcal{C} $ a circle of centre  $O $ and $C $ a point exterior. 
Let $AB $ diameter on the circle. Show that the centers of the circles circumscribed of the triangles $\triangle ABC $ is a line perpendicular on $OC $.
 A: Using coordinates . . .

Let $O=(0,0)$.

Without loss of generality, assume the circle is the standard unit circle, and let $C=(c,0)$, where $c > 1$.

Let the diameter $AB$ be given by $A=(\cos(t),\sin(t))$, and $B=-A$.

We must have $\sin(t)\ne 0$, else triangle $ABC$ is degenerate.

Without loss of generality, assume $0 < t < \pi$.

Let the circumcenter of triangle $ABC$ be $(x,y)$. 

We proceed to solve for $x,y$, in terms of $t$.

Letting $M,N$ be the midpoints of $AC,BC$ respectively, we get
\begin{align*}
M&=\Bigl({\small{\frac{c+\cos(t)}{2}}},{\small{\frac{\sin(t)}{2}}}\Bigr)\\[4pt]
N&=\Bigl({\small{\frac{c-\cos(t)}{2}}},-{\small{\frac{\sin(t)}{2}}}\Bigr)
\end{align*}
$P$ is uniquely determined by the conditions
$$
\begin{cases}
PM\,\bot\,AC\\[4pt]
PN\,\bot\,BC\\
\end{cases}
$$
or equivalently,
$$
\begin{cases}
\vec{PM}\cdot\vec{AC}=0\\[4pt]
\vec{PN}\cdot\vec{BC}=0\\
\end{cases}
$$
which yields the system of equations
$$
\begin{cases}
(2c-2\cos(t))x-(2\sin(t))y=c^2-1\\[4pt]
(2c+2\cos(t))x+(2\sin(t))y=c^2-1\\
\end{cases}
$$
Solving the system, we get
$$
\begin{cases}
x=\frac{c^2-1}{2c}\\[4pt]
y=-x\cot(t)
\end{cases}
$$
so for $t\in (0,\pi)$, $x$ is a positive constant, and $y$ is a continuous, strictly increasing function of $t$, with range $(-\infty,\infty)$.

Since the locus of $P$ is a vertical line, it's perpendicular to $OC$.
