Given the circle, midpoint $O(0,0)$ and radius $r$; choose $A(2a,2b)$. The perpendicular bisector of the segment $OA\ :\ y-a=-\frac{a}{b}(x-a)$ intersects the circle $x^{2}+y^{2}=r^{2}$ in the points $B(a-bw,b+aw)$ and $C(a+bw,b-aw)$ with $w=\sqrt{\frac{r^{2}}{a^{2}+b^{2}}-1}$. The circle $(OBC)\ :\ x^{2}+y^{2}-\frac{ar^{2}}{a^{2}+b^{2}}x-\frac{br^{2}}{a^{2}+b^{2}}y=0$ has midpoint $(\frac{ar^{2}}{2(a^{2}+b^{2})},\frac{br^{2}}{2(a^{2}+b^{2})}\ )$ and radius $\frac{r^{2}}{2\sqrt{a^{2}+b^{2}}}$. Point $D(-\frac{4a^{3}+4a^{2}bw+a(4b^{2}-3r^{2})+bw(4b^{2}-r^{2})}{r^{2}},\frac{4a^{3}w-4a^{2}b+aw(4b^{2}-r^{2})-b(4b^{2}-3r^{2})}{r^{2}}\ )$. Point $E(-\frac{4a^{3}-4a^{2}bw+a(4b^{2}-3r^{2})-bw(4b^{2}-r^{2})}{r^{2}},-\frac{4a^{3}w+4a^{2}b+aw(4b^{2}-r^{2})+b(4b^{2}-3r^{2})}{r^{2}}\ )$. The circle $(ADE)\ :\ x^{2}+y^{2}-\frac{ar^{2}}{a^{2}+b^{2}}x-\frac{br^{2}}{a^{2}+b^{2}}y=2(2a^{2}+2b^{2}-r^{2})$ has the same midpoint and radius $\frac{r^{2}-4(a^{2}+b^{2})}{2\sqrt{a^{2}+b^{2}}}$.