So far no one has posted a solution. So I doubt my solution is correct. Here we go. Let $O$ be the center of $C_1$ (and $C_2$'s as well), the line $AB$ intersects $C_2$ at the points $D$ and $C$, where $A$ is located between the points $D$ and $B$, and finally, the line $XY$ intersects $C_2$ for the second time at the point $Z$. Since the points $X$ and $Y$ are located on the different sides of the point $B$, the point $Z$ would be between the points $B$ and $Y$. Observe that $OA=OB=\frac{1}{2}$, $OC=OD=\frac{k}{2}$ and $DA=BC=\frac{k-1}{2}$. Since $AY=k$ and $OZ=\frac{k}{2}$, then $Z$ is the midpoint of the segment $BY$. So, $BZ=ZY$. From the power of a point in $C_2$ and AM-GM inequality we have $\frac{k+1}{2}\cdot \frac{k-1}{2}=DB\cdot BC=XB\cdot BZ=XB\cdot BY/2\le((\frac{XB+BY}{2})^2)/2$ or $XY\geq\sqrt{2k^2-2}$. Equality occurs iff $XB/BY=1$. Any correction or addition is welcome. Thanks to Dawut.