The circumscribed centers are on a line

Question

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 $.

Answer 1

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 $OC$.