$\definecolor{A}{RGB}{0,148,79}\color{A}\fbox{Coordinates}$ Let $\definecolor{A}{RGB}{0,148,79}\color{A}a,b,c$ be complex numbers lying on unit circle centered at $\definecolor{A}{RGB}{0,148,79}\color{A}0$, representing vertices of $\definecolor{A}{RGB}{0,148,79}\color{A}\triangle ABC.$ $\definecolor{A}{RGB}{190,0,60}\color{A}\fbox{Lemma 1.}$ In non-isosceles $\definecolor{A}{RGB}{80,0,200}\color{A}\triangle ABC$ the center of nine-point circle of triangle $\definecolor{A}{RGB}{80,0,200}\color{A} AHO$ is given by complex number $$\definecolor{A}{RGB}{80,0,200}\color{A} \frac{bc(2a+b+c)-a^3}{2(bc-a^2)}.$$$\definecolor{A}{RGB}{190,0,60}\color{A}\fbox{Proof}$ Triangle is non-isosceles, thus $|AB|\neq |AC|\iff bc-a^2\neq0$. The point is equidistant from midpoints of $\triangle AHO$: $$\left|\frac{bc(2a+b+c)-a^3}{2 (bc-a^2)}-\frac{a}2\right|=\left|-\frac{b c (a + b + c)}{2 (a^2 - b c)} \right|=\left|\frac{(a + b + c)}{2 (a^2 - b c)} \right|,$$$$\left|\frac{bc(2a+b+c)-a^3}{2 (bc-a^2)}-\frac{2a+b+c}{2}\right|=\left| -\frac{a^2 (a + b + c)}{2 (a^2 - b c)}\right|=\left|\frac{(a + b + c)}{2 (a^2 - b c)} \right|,$$$$\left|\frac{bc(2a+b+c)-a^3}{2 (bc-a^2)}-\frac{a+b+c}2\right|=\left| -\frac{a (a b + a c + b c)}{2 (a^2 - b c)}\right|=\left|\frac{(a + b + c)}{2 (a^2 - b c)} \right|.\blacksquare$$$\definecolor{A}{RGB}{190,0,60}\color{A}\fbox{Lemma 2.}$ In non-isosceles $\definecolor{A}{RGB}{80,0,200}\color{A}\triangle ABC$ the point $\definecolor{A}{RGB}{80,0,200}\color{A} X$ given by complex number $$\definecolor{A}{RGB}{80,0,200}\color{A}x=\frac{a^2+b^2+c^2+ab+bc+ca}{2(a+b+c)}$$and lies on nine point circles of $\definecolor{A}{RGB}{80,0,200}\color{A}\triangle ABC, \triangle AHO,\triangle BHO,\triangle CHO.$ $\definecolor{A}{RGB}{190,0,60}\color{A}\fbox{Proof}$ $\triangle ABC$ is non-isosceles thus it's orthocenter and circumcenter are different and $a+b+c\neq0$. Because this is symmetric expression in $(a,b,c)$ it is sufficient to prove it only for $\triangle ABC, \triangle AHO$. From$\definecolor{A}{RGB}{190,0,60}\color{A}\text{ lemma } 1.$ we have $$\left|\frac{bc(2a+b+c)-a^3}{2 (bc-a^2)}-x\right|=\left|\frac{-(a b + a c + b c)^2}{2 (a + b + c) (a^2 - b c)} \right|=\left|\frac{a + b + c}{2 (a^2 - b c)} \right|$$and because $\frac{a+b+c}{2}$ is the nine-point circle center in $\triangle ABC$: $$\left|x-\frac{a+b+c}2\right|=\left|\frac{ab+bc+ca}{2(a+b+c)}\right|=\frac12.\blacksquare$$$\definecolor{A}{RGB}{20,180,255}\color{A}\fbox{Solution}$ Take $X$ as in $\definecolor{A}{RGB}{190,0,60}\color{A}\text{ lemma } 2.$ Then both center of nine point circle of $\triangle ABC$ and point $X$ are those common points. Of course they are not equal points, since $X$ lies on the nine point circle.$\definecolor{A}{RGB}{20,180,255}\color{A}\blacksquare$ #1757 [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -66.33839095628062, xmax = 56.58844031224268, ymin = -47.60935818839934, ymax = 47.249814237854906; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ffqqff = rgb(1,0,1); draw((-35.254648096043326,34.68121175289185)--(-60.83960358295784,-19.03714749921037)--(52.754207825176024,-19.417059911946613)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-35.254648096043326,34.68121175289185)--(-60.83960358295784,-19.03714749921037), linewidth(0.69)); draw((-60.83960358295784,-19.03714749921037)--(52.754207825176024,-19.417059911946613), linewidth(0.69)); draw((52.754207825176024,-19.417059911946613)--(-35.254648096043326,34.68121175289185), linewidth(0.69)); /* special point *//* special point */ draw((-60.83960358295784,-19.03714749921037)--(-35.29532395251842,22.519130666841747), linewidth(0.69) + green); draw((-35.254648096043326,34.68121175289185)--(-35.29532395251842,22.519130666841747), linewidth(0.69) + red); draw((-35.29532395251842,22.519130666841747)--(-4.022359950653367,-13.14606316255344), linewidth(0.69)); draw((-60.83960358295784,-19.03714749921037)--(-4.022359950653367,-13.14606316255344), linewidth(0.69) + green); draw((-4.022359950653367,-13.14606316255344)--(-35.254648096043326,34.68121175289185), linewidth(0.69) + red); draw((-4.022359950653367,-13.14606316255344)--(52.754207825176024,-19.417059911946613), linewidth(0.69) + blue); draw((-35.29532395251842,22.519130666841747)--(52.754207825176024,-19.417059911946613), linewidth(0.69) + blue); /* special point *//* special point *//* special point *//* special point *//* special point *//* special point *//* special point */ draw(circle((-9.107309454595756,-29.860799130032728), 36.1227497196293), linewidth(0.69) + blue); draw(circle((-33.407944803472326,-1.1765440317531455), 14.947023460601832), linewidth(0.69) + green); draw(circle((-41.01139491874429,7.798501254196092), 21.578133838428478), linewidth(0.69) + red); /* special point *//* special point *//* special point *//* special point *//* special point *//* special point */ draw(circle((-19.658841951585895,4.686533752144153), 28.560917565067168), linewidth(0.69) + linetype("4 4")); /* special point */ draw((-41.450858509861334,-13.775157028562006)--(-19.658841951585895,4.686533752144153), linewidth(0.69) + linetype("4 4") + ffqqff); /* dots and labels */ dot((-35.254648096043326,34.68121175289185),linewidth(3pt) + dotstyle); label("$A$", (-35.511538532904815,36.87905215715109), NE * labelscalefactor); dot((-60.83960358295784,-19.03714749921037),linewidth(3pt) + dotstyle); label("$B$", (-63.6297145815814,-25.064013046047346), NE * labelscalefactor); dot((52.754207825176024,-19.417059911946613),linewidth(3pt) + dotstyle); label("$C$", (51.16519775627842,-25.444591471027305), NE * labelscalefactor); label("$H$", (-34.94067089543489,23.08308425162765), NE * labelscalefactor); label("$O$", (-3.63809544083414,-12.596143090243308), NE * labelscalefactor); label("N", (-19.051521652522048,5.576476702549632), NE * labelscalefactor); label("$X$", (-42.17166097005391,-9.025275452773373), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

It is a Special Case of the following Property. Also $O$ can be replaced with any other point in the Original Problem. Property:- Let $P,Q$ be two points on a Rectangular Hyperbola $\mathcal H$ passing through $A,B,C$. Then the Nine Point Circles of $\Delta APQ,\Delta BPQ,\Delta CPQ$ are coaxial. Let $O$ be the Center of $\mathcal H$. It's well known that the Nine Point Circles of $\Delta APQ,\Delta BPQ,\Delta CPQ$ passes through $O$ (See Lemma 2 here). Also the Midpoint of $PQ$ (define it as $K$) lies on all those three circles. Hence, the Nine Point Circles of $\Delta APQ,\Delta BPQ,\Delta CPQ$ are coaxial with radical axis of $\overline{OK}$. $\blacksquare$

Dear Mathinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Boubals.pdf p. 7-10... Sincerely Jean-Louis