Claim Two circles $\omega_1$ and $\omega_2$ are tangent to each other at $T$. If a line intersects $\omega_1$ at $A$ and $B$ and tangent to $\omega_2$ at $X$, then second intersection of $TX$ and $\omega_1$ $S$ is the midpoint of one of the arcs $\overarc{AB}$. Proof If we can prove that the tangent from $S$ to $\omega_1$ is parallel to $AB$, then we are done. Also $AB$ is the tangent to $\omega_2$ at $X$ so if $O_1S\parallel O_2X$ then we are also done. But showing this easy because we know that $O_1, T, O_2$ and $S,T,X$ collinear. So we have $$\angle{O_1ST}=\angle{O_1TS}=\angle{O_2TX}=\angle{O_2XS}~~\text{and we are done}$$ So we have $\overarc{SCT}=\overarc{TX}$. That is equivalent to $\angle{TCI}=\angle{TYI}$, which implies $C,T,I,Y ~~\text{concyclic}.$ $(a)$ is done.

All angle chases left to readers

You can solve the problem using angle chase only, observing equal angles carefully or just trying random equations is enough to solve the problem Let $O_1,O_2$ be centers of circles $k_1,k_2$ respectively. We have homothety between triangles $O_1TS, O_2TX$ $(\clubsuit)$ which is obvious. Let $Z= O_1O_2 \cap XA$. Claim 1. $S$ is midpoint of arc $AB$ i.e $AS=BS$. By tangent property, we have $ \angle O_2XB=90^{\circ}$. Then $$ \angle O_1ZA = \angle XZO_2 = 90^{\circ}-\angle XO_1Z \stackrel{(\clubsuit)}{=} 90^{\circ} - \angle ZO_2S$$Hence $O_1S \perp AB$. Then it is well known that $AS=BS$. $\square$. Claim 2. $TIYC$ are cyclic. $$ \angle TCI= 180^{\circ} - \angle TCS \Rightarrow TO_1S=360^{\circ}-( 360^{\circ}- 2\angle TCS)= 2 \angle TCS$$But We have $$ 2 \angle TCS \stackrel{(\clubsuit)}{=} 2 \angle XO_2T = \angle IYT$$Hence we have proven our claim. $\square$ Let $D$ be intersection of circle $(TIYC)$ and line $AC$. Claim 3. $ \angle BCI= \angle ABS$. $$ \angle BCI = \angle BCT + \angle TCI = ( \angle BAS- \frac{\angle TO_1S}{2}) + \angle TYI= \angle BAS$$Since we know that $ \frac{ \angle TO_1S}{2} = \angle TYI$. $\square$ Claim 4. $ \angle BCA = 180^{\circ}- 2\angle ABS$ Direct consequence of Claim 1. $\square$ Claim 5. $CI$ is bisector of angle $ \angle ECD$. Direct consequence of Claim 4. and Claim 3. $\square$ Claim 6. $SI$ is tangent to circle $(TIX)$ By Claim 2 and basic angle chase, we can see that $$\angle TIC= \angle TXI$$But this is a property of tangent line, hence conclusion. $\square$ Claim 7. $SB$ is tangent to circle $(BTX)$. Similiar to Claim 6, we have $$ \angle SBT = \angle BXT.$$$\square$ Claim 8. $SB=SI$ Direct consequence of Claim 5,6 since $S$ lies on radical axis of the circles ang $S$ is power point of each circle, hence it is well known that $SB=SI$, consequently $SY=SA$ by claim 1.$\square$ Claim 9. $I$ is excenter of $ABC$. Well known after proving our latter claims, if $A,B,C,S$ are cyclic points such that $AS=BS$, and if we know that $IS=BS=AS$ and $IS$ is angle bisector of external angle of $C$, then $I$ is $A-excenter$ of $ABC$. Which we have shown by our claims, hence we are done! [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(17.36337394623623cm); real labelscalefactor = 0.1; /* changes label-to-point distance */ pen dps = linewidth(0.1) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -8.452434540931394, xmax = 8.910939405304834, ymin = -7.926393976978712, ymax = 0.41421048017120354; /* image dimensions */ pen eqeqeq = rgb(0.8784313725490196,0.8784313725490196,0.8784313725490196); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pair O_1 = (-2.931964724884746,-3.8024524499625474), T = (0.8742639629400576,-3.409431768769399), O_2 = (3.2746912527656806,-3.161570264383272), X = (2.989000375524323,-0.7653509042751026), A = (-6.231485873291579,-1.8646696264546856), B = (-0.18052350870167239,-1.1432395375098683), Y = (3.2621406720134565,-5.574727735834093), C = (0.46688818028571666,-5.560189918664959), I = (3.155480164343957,-3.6966797838417103), D = (1.8990123066977722,-6.350298641735329); /* draw figures */ draw(circle(O_1, 3.8264660040128478), linewidth(1.) + blue); draw(circle(O_2, 2.4131901083619876), linewidth(1.) + blue); draw(circle((1.8689646290206325,-4.711798000093499), 1.638776133458893), linewidth(1.) + red); draw(Y--T, linewidth(1.)); draw(C--Y, linewidth(1.) + linetype("4 4")); draw(T--I, linewidth(1.)); draw(circle((2.174924803378208,-2.2819768347386704), 1.7212998722243344), linewidth(1.) + linetype("4 4") + eqeqeq); draw(C--B, linewidth(1.) + linetype("4 4")); draw(I--Y, linewidth(1.)); draw(O_2--Y, linewidth(1.)); draw(O_2--T, linewidth(1.)); draw(X--T, linewidth(1.)); draw(X--B, linewidth(1.)); draw(B--T, linewidth(1.)); draw(B--(-2.478960038004651,-7.602008849534833), linewidth(1.) + qqwuqq); draw(B--A, linewidth(1.)); draw(A--(-2.478960038004651,-7.602008849534833), linewidth(1.) + qqwuqq); draw(O_1--T, linewidth(1.)); draw(O_1--(-2.478960038004651,-7.602008849534833), linewidth(1.)); draw(X--I, linewidth(1.) + linetype("4 4")); draw(Y--(xmax, -0.005200895880482592*xmax-5.5577616818514635), linewidth(1.)); /* ray */ draw(O_2--X, linewidth(1.)); draw(B--I, linewidth(1.)); draw(circle((1.4979810354523042,-1.740557071626131), 1.7816188541066467), linewidth(1.) + linetype("4 4") + eqeqeq); draw(C--D, linewidth(1.)); draw(T--(-2.478960038004651,-7.602008849534833), linewidth(1.)); draw(circle((-2.478960038004651,-7.602008849534833), 6.855546039900086), linewidth(1.) + linetype("4 4") + qqwuqq); draw((-2.478960038004651,-7.602008849534833)--I, linewidth(1.) + qqwuqq); draw((2.989000375524323,-0.7653509042751024)--(xmax, 0.11922567774782582*xmax-1.1217164998354958), linewidth(1.)); /* ray */ draw(A--I, linewidth(1.) + linetype("4 4")); draw(A--C, linewidth(1.) + linetype("4 4")); /* dots and labels */ dot(O_1,linewidth(4.pt)); label("$O_1$", (-2.8950634140847016,-3.715161846487533), NE * labelscalefactor); dot(T,linewidth(4.pt)); label("$T$", (0.9068696972441321,-3.314958361084484), NE * labelscalefactor); dot(O_2,linewidth(4.pt)); label("$O_2$", (3.308090609662343,-3.069378949587158), NE * labelscalefactor); dot(X,linewidth(4.pt)); label("$X$", (3.026129063128386,-0.6772535709280221), NE * labelscalefactor); dot(A,linewidth(4.pt)); label("$A$", (-6.196742168659741,-1.7960042233047282), NE * labelscalefactor); dot(B,linewidth(4.pt)); label("$B$", (-0.14821221881841495,-1.0683615225719112), NE * labelscalefactor); dot((-2.478960038004651,-7.602008849534833),linewidth(4.pt)); label("$S$", (-2.422095658608387,-7.83543863938711), NE * labelscalefactor); dot(Y,linewidth(4.pt)); label("$Y$", (3.298995075903183,-5.479695395764614), NE * labelscalefactor); dot(C,linewidth(4.pt)); label("$C$", (0.5066662118410971,-5.488790929523775), NE * labelscalefactor); dot(I,linewidth(4.pt)); label("$I$", (3.189848670793264,-3.624206508895931), NE * labelscalefactor); dot(D,linewidth(4.pt)); label("$D$", (1.9346650120291995,-6.280102366570714), NE * labelscalefactor); dot((-2.4789600380046535,-7.6020088495348315),linewidth(4.pt)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Remark: There is a case where Circles touch themselves such that one of them inside of it, still, one may use directed angles to avoid configuration issues and I have a weird mistake in Claim 1 with definition of Z, will fix it later