Circle: finding locus of the chord's mid point

Say $AB$ is a chord and $R$ it's midpoint. Say $P= (c,0)$ and $S= (0,0)$ and $T$ the midpoint of $SP$. Then I claim, that $R$ describe a circle with center in $T$. We use a position vectors to prove it.

Since $\angle APB = 90^{\circ}$ we have $\overrightarrow{PA}\cdot \overrightarrow{PB} =0$ so we have $$(\overrightarrow{A}-\overrightarrow{P})\cdot (\overrightarrow{B}-\overrightarrow{P})= 0\;\;\;\;\Longrightarrow \;\;\;\; \overrightarrow{A}\cdot \overrightarrow{B} -\overrightarrow{P}\cdot (\overrightarrow{A}+\overrightarrow{B}) +\overrightarrow{P}^2 =0$$

Say $S$ is the origin, then $||\overrightarrow{A}|| = ||\overrightarrow{B}|| =r$ and $\overrightarrow{R} = {1\over 2}(\overrightarrow{A}+\overrightarrow{B})$ and $\overrightarrow{T}={1\over 2}(\overrightarrow{S}+\overrightarrow{P}) = {1\over 2}\overrightarrow{P}$

\begin{eqnarray} ||\overrightarrow{RT}||^2 &=& \overrightarrow{RT}^2 = (\overrightarrow{T}-\overrightarrow{R})^2\\ &=& {1\over 4} (\overrightarrow{P}-\overrightarrow{A}-\overrightarrow{B})^2\\ &=& {1\over 4}(\overrightarrow{P}^2+\overrightarrow{A}^2+\overrightarrow{B}^2-2\overrightarrow{P}\cdot (\overrightarrow{A}+\overrightarrow{B})+2\overrightarrow{A}\cdot \overrightarrow{B})\\ &=& {1\over 4}(r^2+r^2-\overrightarrow{P}^2) =const. \end{eqnarray}

Let $\Gamma$ be a circle centered at $O$ with radius $R$ and $P$ a point inside $\Gamma$. Let $ABCD$ a quadrilateral inscribed in $\Gamma$ with orthogonal diagonals $AC,BD$ meeting at $P$. By Thales' theorem the midpoints of $AB,BC,CD,DA$ are the vertices of a rectangle with sides parallel to $AC,BD$. Since $O$ is the circumcenter of $ABCD$, the lines joining $O$ with the previous midpoints are orthogonal to the sides of $ABCD$. By angle chasing those midpoints lie on a circle centered at the midpoint of $OP$, whose diameter equals $\frac{1}{2}\sqrt{BD^2+AC^2}$. Let $u,v$ be the distances of $O$ from $AC,BD$. By the Pythagorean theorem we have $u^2+v^2=OP^2$ and $\left(\frac{1}{2}AC\right)^2+u^2 = OA^2 = R^2$, hence $$\tfrac{1}{2}\sqrt{AC^2+BD^2}=\sqrt{2R^2-OP^2}$$ is constant and the midpoints of $AB,BC,CD,DA$ lie on a fixed circle centered at the midpoint of $OP$. This essentially is the Proposition $11$ from Archimedes' Book of Lemmas, which is used to prove the pizza theorem, too (if you are confident with Italian, you may have a look at page 21 of these notes, too).