Let $\ell$ be tangent at $T$, and let the two tangents from $X$ touch at $A$ and $B$. The key point is to invert around $\Gamma$. Then the inverted picture can be interpreted as follows: Points $\Omega$ and $T$ are fixed. Chord $\overline{AB}$ varies inside $\Gamma$, and $X^\ast Y^\ast Z^\ast$ is the medial triangle of $\triangle TAB$. Thus $(X^\ast Y^\ast Z^\ast)$ is the nine-point circle of $\triangle TAB$, but $X^\ast$ is constrained to lie on some circle $\Omega^\ast$. Since $T$ is fixed, and this is equivalent to the nine-point center of $\triangle TAB$ moving along a fixed circle, say with center $S$ and radius $r$. Then the two circles centered at $S$ with radius $R/2 \pm r$ are tangent to $\Gamma^\ast$. Done.

Incidentally the same problem was discussed in this private forum on 15th February 2018. : https://artofproblemsolving.com/community/c538691h1579050_apmo_and_imotc_marathon Let me copy the problem here Quote: PROBLEM 23

Consider a circle $\omega$ which is tangent to line $\ell$ and any circle $\Omega$. Let the point $V$ is a variable point on $\Omega$ and the tangents to $\omega$ through $V$ meet $\ell$ at $A$, $B$. Consider the circumcircle $\pi(V)$ of $VAB$. Then the circles $\pi(V)$ are tangent to two different circles $\tau_1$, $\tau_2$ while $V$ moving on $\Omega$. Moreover $\tau_1$, $\tau_2$ are both tangent to $\Omega$.

See here

My solution (still posting even though it is almost the same as solutions in the above link) Consider an inversion $\Phi$ around $\Gamma$. $\ell$ touches $\Gamma$ at $L$, and the chord of contact of $X$ wrt $\Gamma$ is $MN$. Then the circumcircle of $XYZ$ goes to the ninepoint circle of $LMN$ under $\Phi$. The midpoint of $MN$ moves on a fixed circle, and $L$ is fixed, so the centroid of $LMN$ moves on a fixed circle. Now the circumcenter is fixed, so the Euler line of $LMN$ gives that the ninepoint center is on a fixed circle, with center at a point $G$. The radius of the ninepoint circle is also fixed, so the ninepoint circle is tangent to two circles centered at $G$ with radii $|R/2 \pm R'|$ where $R$ is the radius of $\Gamma$ and $R'$ is the radius of the locus of the nine-point center of $LMN$. Remark : This kind of homothety can also be used in the solution of USAMO 2015 P2.

Let $\Gamma$ have center $O$ and radius $r$. Suppose that $\Gamma$ is tangent to $\ell$ at $A$ and that the tangents from $X$ to $\Gamma$ meet $\Gamma$ at $B$ and $C$. Let $M$ be the midpoint of $OA$. We invert about $\Gamma$, letting $P'$ denote the image of $P$. We have that $X'$ is the midpoint of $BC$ and that $Y'$ and $Z'$ are the midpoints of $AB$ and $AC$. Hence, the circumcircle of $XYZ$ maps to the nine point circle of $ABC$. Let $N$ be the nine point center of $ABC$. Note that $OMNX'$ is a parallelogram by say, complex numbers, so as $X'$ varies around $\Omega'$, $N$ moves around a fixed circle. The circumcircles of $X'Y'Z'$ are thus circles of radius $\frac r2$ whose centers lie on a fixed circle, from which the problem follows.

This problem was also found by mosquitall sometimes back

Consider an invertion around Γ,thenX',Y',Z'are midpoints of three chords.The radius of circleX'Y'Z'is invariable,their locus are tangent with two certain circles.

[asy][asy] unitsize(0.05inches); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); 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 = -27.00662345680965, xmax = 41.40148523790288, ymin = -61.64123166909056, ymax = 41.19895840196104; /* image dimensions */ /* draw figures */ draw(circle((-0.11550857190276709,2.0557166659751367), 12.657400768257721), linewidth(2)); draw((xmin, -5.3708289230665205*xmin + 70.58438187626555)--(xmax, -5.3708289230665205*xmax + 70.58438187626555), linewidth(2)); /* line */ draw(circle((28.193867797617692,4.773185623160206), 4.78341650017635), linewidth(2)); draw((xmin, -0.2422271534180963*xmin + 15.051175558769039)--(xmax, -0.2422271534180963*xmax + 15.051175558769039), linewidth(2)); /* line */ draw((xmin, 0.6187134482498106*xmin-12.757006821049433)--(xmax, 0.6187134482498106*xmax-12.757006821049433), linewidth(2)); /* line */ draw(circle((20.555364330309267,5.663986053288704), 11.847994325866336), linewidth(2)); draw(circle((10.926002563100468,3.9831024520805025), 6.328700384128869), linewidth(2)); draw(circle((5.655325906081134,2.6096696512432542), 0.9750940642936663), linewidth(2)); draw((2.8642853802655477,14.357367864530245)--(4.70422875582405,2.8246641744144108), linewidth(2)); draw((4.70422875582405,2.8246641744144108)--(6.544172131382551,-8.708039515701422), linewidth(2)); draw((12.328039042650055,4.37259322130732)--(9.436105587016302,-2.1677231471970497), linewidth(2)); draw((9.436105587016302,-2.1677231471970497)--(6.544172131382551,-8.708039515701422), linewidth(2)); draw((7.596162211457803,9.36498054291878)--(12.328039042650055,4.37259322130732), linewidth(2)); draw((7.596162211457803,9.36498054291878)--(2.8642853802655477,14.357367864530245), linewidth(2)); /* dots and labels */ dot((12.328039042650055,4.37259322130732),dotstyle); label("$P$", (13.411167430316334,3.745518891605774), NE * labelscalefactor); dot((32.29976879467116,7.227294507573188),dotstyle); label("$X$", (32.964485165555004,5.626741880710377), NE * labelscalefactor); dot((10.82813772871732,12.428306579922747),linewidth(4pt) + dotstyle); label("$Y$", (9.534707937615957,13.037620322637599), NE * labelscalefactor); dot((13.914483533238435,-4.147928733584274),linewidth(4pt) + dotstyle); label("$Z$", (13.183140401333958,-5.660596053917239), NE * labelscalefactor); dot((2.8642853802655477,14.357367864530245),linewidth(4pt) + dotstyle); label("$E$", (3.092944368863861,14.804829797251013), NE * labelscalefactor); dot((6.544172131382551,-8.708039515701422),linewidth(4pt) + dotstyle); label("$F$", (6.9694038615642375,-9.993109604582385), NE * labelscalefactor); dot((4.70422875582405,2.8246641744144108),dotstyle); label("$X'$", (3.3779781550918297,2.26334320322033), NE * labelscalefactor); dot((7.596162211457803,9.36498054291878),linewidth(4pt) + dotstyle); label("$Y'$", (7.3114444050378005,9.845241916884333), NE * labelscalefactor); dot((9.436105587016302,-2.1677231471970497),linewidth(4pt) + dotstyle); label("$Z'$", (9.42069442312477,-3.437332521339072), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $P$ be the tangency point of $\ell$ and let the tangents from $X$ to $\Gamma$ intersect $\Gamma$ at $E$ and $F$, with $X,Y,Z$ collinear and $W,Z,F$ collinear. We will invert about $\Gamma$. The correct way to think about this is "excircle inversion" for $\triangle XYZ$. As usual, the inverted picture will be quite nice in terms of lengths and homotheties, as is often the case with incircle/excircle inversion. Inverted objects will be denoted with primes. Note that $Y'$ is the midpoint of $PE$, $Z'$ is the midpoint of $PF$, and $X'$ is the midpoint of $EF$. So $(XYZ)$ gets sent to the nine-point circle of $PEF$ (recall that the circumcircle gets sent to the nine point circle of the intouch triangle under incircle inversion). Let $G=\frac{2}{3}X'+\frac{1}{3}P$ be the centroid of $PEF$. Note that it varies on a circle since $X'$ does. We see that a homothety with scale factor $-1/2$ about $G$ sends $\Gamma$ to $\Omega'$, so \[\Omega'=-\frac{1}{2}(\Gamma-G)+G=-\frac{1}{2}\Gamma+\frac{3}{2}G\equiv -\frac{1}{2}\Gamma+K\]where $K$ varies on a circle. So, $\Omega'=\gamma+K$ where $\gamma$ is a fixed circle, and WLOG $K$ varies on the unit circle. This means that $\Omega'$ is tangent to two fixed circles, so inverting back solves the problem. $\blacksquare$

Great problem! Solved with a hint from v_Enhance: Assume all angles are directed. Let $A=\overline{XY}\cap\Gamma,$ let $B=\overline{XZ}\cap\Gamma,$ and let $C=\overline{YZ}\cap\Gamma.$ Invert around $\Gamma.$ Then, $A,B,C$ are fixed and $X^{*},Y^{*},Z^{*}$ are the midpoints of $\overline{AB},\overline{BC},\overline{CA}$ respectively. Note that $(X^{*}Y^{*}Z^{*})$ is the nine-point circle of $\triangle ABC,$ so the radius of $(X^{*}Y^{*}Z^{*})$ is always half the radius of $\Gamma.$ Let this value be $r_1.$ $\textbf{Claim: }$ The line tangent to $(X^{*}Y^{*}Z^{*})$ at $X^{*}$ is parallel to $\overline{YZ}.$ $\textbf{Proof: }$ Let $P$ be a point on the line through $X^{*}$ parallel to $\overline{YZ}.$ We have $$\angle PX^{*}Z^{*}=\angle PX^{*}B+\angle BX^{*}Z^{*}=\angle PX^{*}B+\angle BAC,$$$$\angle X^{*}Y^{*}Z^{*}=\angle Z^{*}BX^{*}=\angle CBA.$$But since $\overline{PX^{*}}$ is parallel to the line tangent to $\Gamma,$ we know $\angle PX^{*}B=\angle CBA-\angle BAC.$ Therefore, $$\angle PX^{*}Z^{*}=\angle X^{*}Y^{*}Z^{*},$$as desired. $\blacksquare$ Now let $U$ be the point on $\Omega$ corresponding to $C$ on $\Gamma,$ let $V$ be the point diametrically opposite $U,$ and let $r_2$ be the radius of $\Omega.$ It is easy to see that $(X^{*}Y^{*}Z^{*})$ is tangent to (1) The circle externally tangent to $\Omega$ at $U$ with radius $r_1-r_2$ (2) The circle internally tangent to $\Omega$ at $V$ with radius $r_1+r_2.$ Thus, $(XYZ)$ is tangent to two fixed circles.

This is not medium geo @v_Enhance. Let $A=XY\cap \Gamma,B=XZ\cap\Gamma,T=YZ\cap\Gamma$. Let $O$ be the center of $\Gamma$. Invert about $\Gamma$ and let objects $\bullet$ go to $\bullet^\ast$. Observe that $X^\ast$ is the midpoint of $AB$, $Y^\ast$ is the midpoint of $AT$, $Z^\ast$ is the midpoint of $BT$. We seek to show that as $X^\ast$ varies, $(X^\ast Y^\ast Z^\ast)$ is tangent to two constant circles. Let $H$ denote the orthocenter of $\triangle TAB$. Since $T$ is the orthocenter of $\triangle HAB$, homothety at $T$ with factor $2$ takes $(X^\ast Y^\ast Z^\ast)$ to $(HAB)$, which is the reflection of $\Gamma$ over line $AB$. Hence, it suffices to show that as $X^\ast$ varies, the reflection of $\Gamma$ over the line through $X^\ast$ perpendicular to $OX^\ast$ is constant to two fixed circles. Now, let $O_2$ denote the reflection of $O$ over $X^\ast$, it is clear that $O_2$ is the center of the described reflection of $\Gamma$. Moreover, it is clear that $O_2$ moves along a circle as $X^\ast$ does, so it suffices to show that as $O_2$ moves along a particular circle, the circle about $O_2$ with fixed radius is always tangent to some two circles. However, this is trivial; let the fixed radius be $R$ and the center of the circle $O_2$ moves along be $O_3$ with radius $r$, then the first circle is the set of points distance $R-r$ away from $O_3$ and the second circle is the set of points distance $R+r$ from $O_3$.

Oops I used geogebra for this Do an inversion centered at $\Gamma$. Let $A, B$ be the points such that $XA$ and $XB$ are tangent to $\Gamma$. From now on, we only refer to the inverted image (so for any point $P$ when we write $P$, we really mean $P^{*}$). Let $O_{1}$ be the center of $\Gamma$, then this means $X$ is the midpoint of $AB$. Furthermore, $\ell$ is the circle tangent to the $\Gamma$ at a point $T$, so then since $Y$ lies on $\Gamma$ and $\angle O_{1}YT = 90$, and also $Y$ lies on $(O_{1}XB)$ so $\angle O_{1}YB = 90$, this means $B, Y, T$ are collinear, and by homothety, $Y$ is the midpoint of $BT$. Similarly, $Z$ is the midpoint of $AT$, so $(XYZ)$ is the nine point circle of $\triangle TAB$. Now, consider $X$ moving around the circle. The radii of the nine point circle is fixed (it's half the radii of $\Gamma$). Furthermore, if $O$ was the center of the nine-point circle, it is well known that $XO || TO_{1}$. Thus, as $X$ moves around a circle, $O$ also moves around a fixed circle. Let $r_{1}$ be the radii of the nine-point circle and $r_{2}$ be the radii of the fixed circle that $O$ moves around, and $P$ the center of the fixed circle. Then, consider the circles with radii $r_{1} -r_{2}, r_{1} + r_{2}$, since for every nine-point circle, when we draw $O$ to $P$ it is a distance of $r_{1} - r_{2}$, and going the other way around it is a distance of $r_{1} + r_{2}$, every nine-point circle are tangent to these two circles, which are fixed. Thus, inverting backs reveals that $(XYZ)$ is always tangent to two fixed circles.

Absolutely beautiful. Let $\ell\cap\Gamma=B$ and the tangency points from $X$ be $E$ and $F$. We can then invert around $\Gamma$, where the images of $X,Y,Z$ are the midpoints of $EF,BF,BE$, and hence $(X^*Y^*Z^*)$ is the nine-pt circle of $BEF$, and $X^*$ varies over $\Omega^*$. Set $\Gamma$ to be the unit circle in the complex plane. The nine-pt center $n$ of $BEF$ is situated at $\frac{b+e+f}{2}$, where $\frac{e+f}{2}=x^*$ varies on a circle and $\frac{b}{2}$ is fixed, and therefore $n$ also varies on a circle. This implies the problem as the nine-pt radius is fixed, and our two desired fixed circles indeed exist with center $o_n=o_\Omega^*+\frac{b}{2}$.

One of my favorites .I have the same solution but I am still posting it. Let the center of $\Gamma$ be $O$. We invert about $O$. It is now easy to see that $(XYZ)$ goes to the nine-point circle of $(TAB)$. $\textbf{Claim:}$ Take any circle $\Gamma$with Radius $R$ and center $N$ and let $K$ be a point on this circle. Draw a circle of radius $r$ with center $K$, then the two circles centered at $O$ with radius $|R-r|$ and $|R+r|$ are tangent to $\Gamma$. $\textbf{Proof:}$ This is quite obvious but I wrote it this way to make things very clear(I know this is weird but this looks more clean to me). $\blacksquare$ Thus, we are done by the claim if we are able to prove that the nine-point center moves on a fixed circle. Now since $X^{*}$ moves along a fixed, so does the centroid. So by Euler's line, this is the same as saying that the nine-point circle moves on a fixed circle $\blacksquare$

Let $W=\ell \cap \Gamma, O$ is the center of $\Gamma.$ Invert wrt $\Gamma$ with same labelings. Let $R,r$ denotes radii of $\Gamma, \Omega.$ Since $XYZ$ is the medial triangle of triangle with circumcircle $\Gamma$ and vertex $W,$ circumcenter of triangle moves along circle $\omega =\mathcal {H}^{3/2}_O\circ \mathcal {H}^{2/3}_W (\Omega).$ Hence $\odot (XYZ)$ tangent to circles concentric with $\omega$ with radii $|R/2\pm r|,$ which are fixed, done.

I'm surprised that nobody has posted the following true (confirmed using geogebra) statement. I'm not sure how to prove this, but inversion at $\Gamma$ is helpful. Let the external common tangents of $\Gamma$ and $\Omega$ intersect $\ell$ at $P$ and $Q$, and let the internal common tangents of $\Gamma$ and $\Omega$ intersect $\ell$ at $R$ and $S$. Let $\omega_1$ be the circle through $P$ and $Q$ internally tangent to $\Omega$. Let $\omega_2$ be the circle through $R$ and $S$ externally tangent to $\Omega$. Then, the circumcircle of $\triangle XYZ$ is internally tangent to $\omega_1$ and $\omega_2$.

Very intuitive-ish geometry problem. Invert about $\Gamma$. Denote by $E$ and $F$ the tangency points of $\overline{XY}$ and $\overline{XZ}$ to $\Gamma$, $K$ and $L$ the images of $Y$ and $Z$, $B$ the tangency point of $\ell$ to $\Gamma$, $P$ the inverse of $X$, and $H$ the orthocenter of triangle $BEF$. Notice foremost that $P$ is the midpoint of $\overline{EF}$ as $P = (OE) \cap (OF)$. $\ell$ inverts to $(OB)$, so by homothety $K, L$ are midpoints of $\overline{BE}, \overline{BF}$. Hence $(PKL)$, the image of $(XYZ)$, is the nine-point circle of $BEF$. It suffices to show that the center $N$ of $(XYZ)$ moves along a circle. To do so, note that $P$ moves along a fixed circle (the image of $\Omega$); As $BH = 2OP$, $H$ moves along a fixed circle by homothety; As $N$ is the midpoint of $\overline{OH}$, $N$ moves along a circle too.

Let $A$ and $B$ be the tangency points from $X$ to $\Gamma$ and $C$ be the tangency point of $\ell$. Invert around $\Gamma$. Then, $X^{\ast}$ is the midpoint of $AB$, $Y^{\ast}$ is the midpoint of $AC$, and $Z^{\ast}$ is the midpoint of $BC$. Thus, $(X^{\ast}Y^{\ast}Z^{\ast})$ is the nine-point circle of $\triangle ABC$. Claim: The nine-point center of $\triangle ABC$ moves along a circle. Let $\Gamma$ be the unit circle. Note that $X^{\ast}=\frac{a+b}{2}$ moves along a circle (the inverse of $\Omega$). Furthermore, the nine-point center is $\frac{a+b+c}{2}$, and since $c$ is fixed, this is just $X^{\ast}$ shifted by $c/2$, so it still moves along a circle. Since the center of $(X^{\ast}Y^{\ast}Z^{\ast})$ moves along a circle, and its radius is fixed (half of $(ABC)$), there are two concentric circles centered at the center of the locus of the nine-point center that are always tangent to it, hence done.

I need to be less scared of P3/6s. We invert about $\Gamma$: inverted problem statement wrote: Fix a circle $\Gamma$ (with center $O_1$ and radius $R$), a point $A$ on $\Gamma$, and another circle $\Omega$ (with center $O_2$ and radius $r$) inside $\Gamma$. The line through a variable point $X$ on $\Omega$ perpendicular to $O_1X$ meets $\Gamma$ at $Y$ and $Z$. Prove that, as $X$ varies over $\Omega$, the nine point circle of $\triangle AYZ$ is tangent to two fixed circles. Note that the NPC of $\triangle XYZ$ always has radius $\frac{1}{2}R$, and since the orthocenter of $\triangle XYZ$ is $A + 2(X-O_1)$, it is centered at $\frac{1}{2}A - \frac{1}{2}O_1 + X$, which moves on a circle with radius $r$ centered at $\frac{1}{2}A - \frac{1}{2}O_1 + O_2$. Hence, it is clearly tangent to the circles centered at $\frac{1}{2}A - \frac{1}{2}O_1 + O_2$ with radii $\left|r\pm \frac{1}{2}R\right|$.

ah yes, so basically inverting is like rolling a die and getting a problem statement, then choosing to reroll the die into a different number that may or may not be easier to use. However this die is infinite so you can only take your chances. Refer to diagram. Invert over $\Gamma$ with radius R, s.t. the inverse of (XYZ) is the ninepointcirc of JKL, and since F moves along the circle inverse of $\Omega$, the centroid of JKL moves along a circle. Since the circumcenter A is fixed, the ninepointcenter of JKL moves along a circle $\omega$ with center W and radius r. Indeed, then the ninepointcirc of JKL is always tangent to the two circles centered at W with radius $R\pm\frac r2$, and inverting back, the inverses of those two circles are then tangent to the original (XYZ).

Let $O$ be center of $\Gamma$. Consider inversion around $\Gamma$. Then $\Omega^*$ lies inside $\Gamma$. Let $X$ be a variable point of $\Omega^*$. Let $l$ be line passing through $X$ and perpendicular to $OX$. Let $X_1, X_2$ be intersection of $l, \Gamma$ and let $Y, Z$ are midpoints of $AX_1, AX_2$, respectively. Then we'll prove that the following statement: As $X$ varies over $\Omega^*$, $(XYZ)$ is tangent to two fixed circle. Note that $AYXZ$ is parallelogram. Thus the the radius of $(XYZ)$ is same as the radius of $(AYZ)$. Since $\angle{AZO} + \angle{AYO} = 90^{\circ} + 90^{\circ} = 180^{\circ}$, so the radius of $(AYZ)$ is $\frac{AO}{2}$. Therefore as $X$ varies over $\Omega^*$, the radius of $(XYZ)$ is constant. Let $O_X$ be center of $(XYZ)$. Let $R$ be radius of $\Omega^*$. Claim: As $X$ varies over $\Omega^*$, $O_X$ moves along a circle with radius $R$. Proof: Note that $OZ \perp AX_2$ and $OY \perp AX_1$. Thus $OZ \perp XY$ and $OY \perp XZ$. Hence $O$ is orthocenter of $XYZ$. Let $G$ be centroid of $AX_1X_2$. Then $G$ becomes centroid of $XYZ$. Therefore $O_X, G, O$ are collinear. Since $A, G, X$ are collinear and $\frac{AX}{AG} = \frac{3}{2}$. Since $X$ moves along $\Omega^*$, so $G$ moves along a circle. Note that $\frac{OG}{OO_X} = \frac{2}{3}$ and since $G$ moves along a circle, so $O_X$ moves along a circle with radius $R$. $\blacksquare$ As $X$ varies $\Omega^*$, the radius of $(XYZ)$ is constant and $O_X$ moves along a circle, so $(XYZ)$ moves along a circle. Therefore as $X$ moves along a circle, $(XYZ)$ tangent to two fixed circles with radius $\frac{AO}{2} - R$ and $\frac{AO}{2} + R$. $\blacksquare$

Invert about $\Gamma.$ Then $\ell$ becomes a circle tangent to $\Gamma$ at a point $P$ and with half the radius, and $\Omega$ becomes a circle inside $\ell.$ We get that the tangents to $\Gamma$ invert to circles through $X,O$ internally tangent to $\Gamma$ at some points $A,B$ and the circumcircle of $XYZ$ inverts to a circle through $X$ and the intersections $Y,Z$ of the circles through $X,O$ and the circle $\ell.$ We may assume that $A,Y,O,X$ are concyclic and $B,Z,O,X$ are concyclic. Then, we can see that $AO,BO,PO$ are diameters of circles $AYOX,BZOX,OYPZ$ respectively, so $\angle PYO=\angle AYO=90$ and $A,Y,P$ are collinear. Furthermore, since a homothety at $P$ with scale factor $2$ sends circle $OYPZ$ to $\Gamma$ we get that this must send $Y$ to $A,$ so $Y$ is the midpoint of $AP.$ Similarly, $Z$ is the midpoint of $BP,$ and for the same reason $X$ is the midpoint of $AB.$ Therefore, circle $XYZ$ is the nine-point circle of $ABP.$ Next, we will show that the nine-point center of $ABP$ varies on a circle as $X$ varies on $\Omega.$ We do this with complex numbers, letting $\Gamma$ be the unit circle. We get $N_9=\frac{a+b+p}{2}=x+\frac{p}{2},$ and since $p$ is fixed, our claim is proved. From here, notice that the radius of the nine-point circle is also fixed, since it is half of the radius $R$ of $\Gamma.$ To finish, let $K$ be the center of the circle containing the nine-point centers, and suppose this circle has radius $r.$ Then, the nine-point circle is simply tangent to the circles centered at $K$ with radii $\frac{R}{2} \pm r.$ This always works, since $r$ is also the radius of $\Omega,$ and since $\Omega$ is inside $\ell$ we find that $\frac{R}{2}>r.$

We invert around $\Gamma$. Let $XY$ and $XZ$ be tangent to $\Gamma$ at $A$ and $B$ respectively. Also, let $\Gamma$ be tangent to $\ell$ at $C$ We can see that by common inversion formulas we have that $XYZ$ is the medial triangle of $ABC$. We can obviously thus see that since the locus of the nine-point centers is a circle that we can just choose the inner circle for this and the outer circle. $\blacksquare$

Suppose that lines $XY$, $XZ$, and $YZ$ touch $\Gamma$ at , , and , respectively. Suppose that , , and are the inverses of $X$, $Y$, and $Z$, respectively, with respect to $\Gamma$, which are the midpoints of , , and , respectively. Then, the circumcircle of $XYZ$ maps to the nine-point circle of under this inversion. Since moves along a fixed circle and $T$ is fixed, the centroid of moves along a fixed circle. Since the circumcenter of is fixed, its nine-point center moves along a fixed circle as well. The nine-point circle has constant radius, so it is always tangent to two fixed circles. We finish by undoing our inversion. $\square$

Let the tangents from $X$ meet $\Gamma$ at $P$ and $Q$, and $T = \ell \cap \Gamma$. Inverting about $\Gamma$, we have: $\ell$ is mapped to $(OT)$. $X$ is mapped to the midpoint of chord $PQ$. $Y \mapsto (OP) \cap (OT)$, $Z \mapsto (OQ) \cap (OT)$, and thus $Y$, $Z$ are the midpoints of $TP$, $TQ$. Thus $(XYZ)$ is now the nine-point circle of $\triangle TPQ$, so it has fixed radius $\frac R2$, where $R$ is the radius of $\Gamma$. Note that $N$, the center of $(XYZ)$, is \[\frac{t+p+q}{2} = x + \frac t2,\]so its locus is a circle with fixed radius $r$, the radius of $\Omega$ (after inversion). Hence we take the two circles centered at locus center with radii $\frac R2 - r$ and $\frac R2 + r$. $\blacksquare$

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Let $A$ be the center of $\Gamma$. Consider an inversion around $\Gamma$, let $X,Y,Z$ invert to $E',I',H'$, We show that $E'$ is the orthocenter of $\triangle AI'H'$. Now let $K$ be the center of $(AI'H')$ and let $J$ be the center of $(E'H'I')$ then the locus $J$ is circle with same radius as $\Omega'$, this is since $AKE'J$ is a parallelogram (can be proves easily). Now to construct these two fixed circles. Let $r$ denote the radius of $\Omega'$, and let $R$ be the radius of $\Gamma$. Circles with radius $\frac{R}{2}+r$ and $\frac{R}{2}-r$ centered at $P$, where $P$ is the center of locus (circle) of $J$. We see that these two circles are tangent to $(E'H'I')$, hence we are done.

Solved with ohiorizzler1434, even though solutions might still be similar to above solutions. Invert around $\Gamma$. Let $O$ be its circumcentre. If $\ell$ is tangent to $\Gamma$ at $A$ then $A$ is fixed, whilst $\Omega\to\Omega’$, which by abuse of notation we rename $\Omega$ again. Now $X$ is sent to some point on $\Omega$ which we don’t care about, all we care is the fact that it lies on $\Omega$. Now it can be seen that if the tangency points from $X$ to $\Gamma$ before inversion are $X_1$, $X_2$ then they are fixed and $\measuredangle X_1XO=\measuredangle X_2XO=90^{\circ}$ after inversion (well-known). hence $X_1X_2$ is the chord on $\Gamma$ such that $X$ is its midpoint. Now $Y$ and $Z$ invert to $X_1A$’s and $X_2A$’s midpoints (we don’t care which). So now $(XYZ)$ is literally just the nine-point circle of $\triangle AX_1X_2$ with centre $N_9$. Its radius $r$ is fixed as it is just half of the radius of $\Gamma$ which itself is fixed. Now we claim that $N_9$ in fact lies on a fixed circle $\gamma$, which suffices as then if the radius of the fixed circle is $R$, the tangent circles can thus be described as the two circles concentric with $\gamma$ having radii $R-r$ and $R+r$. Since the circumcentre of $\triangle AX_1X_2$ is literally just $O$ which is fixed we can just dilate from $O$ by a factor of $2$ to get the STP that the orthocentre $H$ of $\triangle AX_1X_2$ lies on a fixed circle. However note that $AH=2XO$ which is well known. Hence the curve $H$ lies on can just be written as $2(\Omega+A-O-A)+A=2(\Omega-O)+A$ which is just a circle (here all these operations are just your standard $\mathbb{R}^2$ vector space operations). We are thus done.

took half the OTIS hints but made it in the end

Lovely problem! Invert with respect to $\Gamma$. Let $XY$ and $XZ$ be tangent to $\Gamma$ at $P$ and $Q,$ respectively, and let $\ell$ touch $\Gamma$ at $T.$ Under inversion, $Y$ is sent to the midpoint of $TP,$ and $Z$ is sent to the midpoint of $TQ.$ Similarly, $X$ is sent to the midpoint of $PQ.$ Finally, the inverse of $X$ lies on a fixed circle. Therefore, our translated problem is as follows: Inverted problem wrote: Let $\Gamma$ and $\Omega^{\ast}$ be two fixed circles, with $\Omega^{\ast}$ inside $\Gamma.$ Let $T$ be a fixed point on $\Omega,$ and let $X^{\ast}$ be a point varying on $\Omega^{\ast}.$ Points $P^{\ast}$ and $Q^{\ast}$ are on $\Gamma$ so that $X^{\ast}$ is the midpoint of $P^{\ast}Q^{\ast}.$ Prove that as $X^{\ast}$ varies on $\Omega^{\ast},$ the nine-point circle of $\triangle TP^{\ast}Q^{\ast}$ is tangent to two fixed circles. We now employ complex numbers, setting the unit circle to be $\Gamma.$ For any point $A^{\ast},$ let $a$ be the complex number corresponding to it. Then the orthocenter of $\triangle TP^{\ast} Q^{\ast}$ is given by $h = p + q + t = 2x + t$ since $X^{\ast}$ is the midpoint of $P^{\ast}Q^{\ast}.$ Therefore, the nine-point circle is a circle of radius $\frac{1}{2}$ centered at $x + \frac{t}{2}.$ Since $x$ lies on a fixed circle, so does the center of our nine-point circle. We must show that as $x$ varies along this circle, the circle of radius $\frac{1}{2}$ is tangent to two fixed circles. This is clear; just take the circle centered at $\frac{t}{2}$ with radius $|R \pm 0.5|,$ where $R$ is the radius of $\Omega^{\ast}.$ Thus we are done.