Given the fixed circle, radius $r$ and midpoint $O$. On this circle, a fixed point $C(0,-r)$ and two points $A(r\cos \alpha,r\sin \alpha),B(r\cos \beta,r\sin \beta)$. Slope of the line $AC\ :\ m_{AC}=\frac{\sin \alpha+1}{\cos \alpha}$, slope of the line $l \bot AC\ :\ m_{l}=\frac{\sin \alpha-1}{\cos \alpha}$. Equation of $l$, through $B\ :\ y-r\sin \beta=\frac{\sin \alpha-1}{\cos \alpha} \cdot (x-r\cos \beta)$. Slope of the line $BC\ :\ m_{BC}=\frac{\sin \beta+1}{\cos \beta}$, slope of the line $l' \bot BC\ :\ m_{l'}=\frac{\sin \beta-1}{\cos \beta}$. Equation of $l'$, through $A\ :\ y-r\sin \alpha=\frac{\sin \beta-1}{\cos \beta} \cdot (x-r\cos \alpha)$. Crossing point $H = l \cap l'\ :\ H(r(\cos \alpha+\cos \beta),r(\sin \alpha+\sin \beta-1))$. This point $H$ lies on a circle, midpoint $C$ and radius $s$, giving a condition between $\alpha$ and $\beta$, namely $\cos(\alpha-\beta)=\frac{s^{2}}{2r^{2}}-1$. $AC$ and $l$ cut in $B'(\frac{r}{2}(\cos \alpha+\cos \beta-\sin \alpha\cos \beta+\sin \beta\cos \alpha),\frac{r}{2}(\cos \alpha\cos \beta+\sin \alpha\sin \beta+\sin \alpha+\sin \beta-1))$. $BC$ and $l'$ cut in $A'(\frac{r}{2}(\cos \alpha+\cos \beta+\sin \alpha\cos \beta-\sin \beta\cos \alpha),\frac{r}{2}(\cos \alpha\cos \beta+\sin \alpha\sin \beta+\sin \alpha+\sin \beta-1))$. We see $y_{A'}=y_{B'}$. Midpoint $M$ of $A'B'\ :\ M(\frac{r}{2}(\cos \alpha+\cos \beta),\frac{r}{2}(\cos \alpha\cos \beta+\sin \alpha\sin \beta+\sin \alpha+\sin \beta-1))$. Eliminating the parameter $\alpha$, we find the locus, the circle $x^{2}+[y-(\frac{s^{2}}{4r}-r)]^{2}=\frac{s^{2}}{4}$. Drawing the picture, we note, the real locus is half this circle.