Let $M$ be the midpoint of $AC$. Lemma : $M$ is the incentre of $\triangle XYZ$. Proof : Consider quad. $A_{1}MAX$. $\angle A_{1}MC = 180 - 2C = \angle A_{1}XA$. So quad. $A_{1}MAX$ is cyclic. But $MA = MA_{1}$. So $\angle AXM = \angle A_{1}XM$ which means that $XM$ is the $\angle$ bisector of $\angle ZXY$. Similarly $YM$ is the $\angle$ bisector of $\angle ZYX$. $\therefore M$ is the incentre of $\triangle XYZ$. So $\odot ABC$ is tangent to $XZ$ and $YZ$ at $A$ and $C$ respectively and midpoint of $AC$ is the incentre of $\triangle XYZ$. This means that $\odot ABC$ is the mixtillinear incircle of $\triangle XYZ$ opposite to $Z$. $\implies \odot ABC$ and $\odot XYZ$ are tangent to each other as desired.

Let $H$ be the orthocenter of $\triangle ABC$ and $H_A, H_B, H_C$ are reflections of $H$ across $BC, CA, AB$ respectively. Observe that $\triangle XZY$ and $\triangle H_AH_BH_C$ are homothetic. Moreover, by APMO 2008 P3, lines $XH_C$ and $YH_A$ meet at $T\in\odot(ABC)$. Hence the circles are tangent at $T$.