This was easier than I thought (hopefully I didnt make a mistake). Suppose that $AB > BC$. Claim 1. $Y$ lies on the incircle $\iff b^2=r_b^2-2rr_b$ Proof: Notice that if $Y$ lies on the incircle, we must have $IY=r$ and since $IC_1\parallel JC_2$ we get $IC_2=r+r_B$. Now from recall that $C_1C_2=b$ so by Pythagoras in $\triangle IC_1C_2$ we get that $IY=r \iff b^2=r_b^2-2rr_b$. If we do the same for point $Z$ we get the same length condition so forget about $Z$. Claim 2. $X$ lies on the incircle $\iff (a-c)^2=r_b^2-2rr_b$ Proof: Since $X$ lies on the incircle we must have $IX=r$ and from $IB_1\parallel JB_2$ it follows that $B_2I=r_b-r$. Now apply Pythagoras theorem in $\triangle IB_1B_2$ and note that $B_1B_2 = c-a$ so $(a-c)^2=r_b^2-2rr_b$ (this can be done the other way around too). Claim 3. $b^2-(a-c)^2=4rr_b$ Proof: it is well know that $r_b=\frac{S}{p-b}$ where $S$ is the area of $\triangle ABC$ and also $rp=S$. We have: $$b^2-(a-c)^2=(b-a+c)(b+a-c)=4(p-a)(p-c)=4\frac{S^2}{p(p-b)}=4rr_b$$ Now by combining the claims we get the desired conclusion. $\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -4.321357124490667, xmax = 34.37305935103295, ymin = -16.990293689875713, ymax = 8.97413454331688; /* image dimensions */ pen qqzzcc = rgb(0.,0.6,0.8); pen ffwwqq = rgb(1.,0.4,0.); pen qqzzqq = rgb(0.,0.6,0.); draw((18.,-6.)--(9.269166453569008,4.22635338067595)--(8.,-6.)--cycle, linewidth(0.8) + qqzzcc); /* draw figures */ draw((18.,-6.)--(9.269166453569008,4.22635338067595), linewidth(0.8) + qqzzcc); draw((9.269166453569008,4.22635338067595)--(8.,-6.), linewidth(0.8) + qqzzcc); draw((8.,-6.)--(18.,-6.), linewidth(0.8) + qqzzcc); draw(circle((13.,-1.4286039736161944), 6.774781297579843), linewidth(0.8)); draw(circle((11.429202882937206,-2.970078183861028), 3.029921816138972), linewidth(0.8)); draw(circle((14.570797117062794,-13.436692358531046), 7.436692358531046), linewidth(0.8)); draw((11.429202882937206,-2.970078183861028)--(20.226602848101475,-8.608001566174991), linewidth(0.8) + ffwwqq); draw((13.73354027436116,-1.002730883966988)--(14.570797117062794,-13.436692358531046), linewidth(0.8) + ffwwqq); draw((11.429202882937206,-2.970078183861028)--(13.73354027436116,-1.002730883966988), linewidth(0.8)); draw((14.570797117062794,-13.436692358531046)--(20.226602848101475,-8.608001566174991), linewidth(0.8)); draw((18.,-6.)--(20.226602848101475,-8.608001566174991), linewidth(0.8)); draw((8.,-6.)--(7.1907239065885795,-12.520770613252042), linewidth(0.8)); draw((9.26916645356901,-0.8868233096620192)--(14.570797117062794,-13.436692358531046), linewidth(0.8) + qqzzqq); draw((14.570797117062794,-6.)--(9.26916645356901,-0.8868233096620192), linewidth(0.8) + qqzzqq); draw((11.429202882937206,-2.970078183861028)--(11.429202882937206,-6.), linewidth(0.8)); draw((14.570797117062794,-13.436692358531046)--(14.570797117062794,-6.), linewidth(0.8)); /* dots and labels */ dot((18.,-6.),dotstyle); label("$A$", (18.611908868987747,-6.214313428603501), NE * labelscalefactor); dot((9.269166453569008,4.22635338067595),dotstyle); label("$B$", (8.438304750106845,4.107658851609926), NE * labelscalefactor); dot((8.,-6.),dotstyle); label("$C$", (7.240380356246128,-6.329744280334583), NE * labelscalefactor); dot((11.429202882937206,-2.970078183861028),linewidth(4.pt) + dotstyle); label("$I$", (11.254036138089957,-2.5469502019529184), NE * labelscalefactor); dot((14.570797117062794,-13.436692358531046),linewidth(4.pt) + dotstyle); label("$J$", (14.599441158339576,-13.092068759762968), NE * labelscalefactor); dot((14.570797117062794,-6.),linewidth(4.pt) + dotstyle); label("$B_2$", (14.399441158339576,-5.973660693136512), NE * labelscalefactor); dot((11.429202882937206,-6.),linewidth(4.pt) + dotstyle); label("$B_1$", (11.554036138089957,-5.773660693136512), NE * labelscalefactor); dot((8.422349351207295,-2.5969052338254373),linewidth(4.pt) + dotstyle); label("$A_1$", (8.527325646906363,-2.3611929824883426), NE * labelscalefactor); dot((13.73354027436116,-1.002730883966988),linewidth(4.pt) + dotstyle); label("$C_1$", (13.838905822610908,-0.7588168400970284), NE * labelscalefactor); dot((20.226602848101475,-8.608001566174991),linewidth(4.pt) + dotstyle); label("$C_2$", (20.337431288975687,-8.384940332589025), NE * labelscalefactor); dot((7.1907239065885795,-12.520770613252042),linewidth(4.pt) + dotstyle); label("$A_2$", (7.310706723979623,-12.272186159501288), NE * labelscalefactor); dot((13.975913095671787,-4.602169155421846),linewidth(4.pt) + dotstyle); label("$Y$", (14.373031203408013,-4.675736299275797), NE * labelscalefactor); dot((9.26916645356901,-0.8868233096620192),linewidth(4.pt) + dotstyle); label("$X$", (9.061451027703468,-0.5511014142314876), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

As above the conditions re-write to $IB_2=r_b-r$, $IC_2=r_b+r$, and $IA_2=r_b+r$. Applying Pythagoras Theorem to $IA_1A_2$, $IB_1B_2$, and $IC_1C_2$ it suffices to show that any one of $$r^2+(a-c)^2=(r_b-r)^2$$$$r^2+b^2=(r_b+r)^2$$$$r^2+b^2=(r_b+r)^2$$implies the other two. The last two are equivalent so it suffices to show that $$(r^2+b^2)-(r^2+(a-c)^2)=(r_b+r)^2-(r_b-r)^2 \iff 4rr_b=\frac{4\Delta^2}{s(s-b)}=4 (s-a)(s-c)=(a+b-c)(-a+b+c)=b^2-(a-c)^2$$Implying the result as desired.