Three circles $C_i$ are given in the plane: $C_1$ has diameter $AB$ of length $1$; $C_2$ is concentric and has diameter $k$ ($1 < k < 3$); $C_3$ has center $A$ and diameter $2k$.  We regard $k$ as fixed.  Now consider all straight line segments $XY$ which have one endpoint $X$ on $C_2$, one endpoint $Y$ on $C_3$, and contain the point $B$.  For what ratio $XB/BY$ will the segment $XY$ have minimal length?
1. **Define the centers and radii of the circles:**
   - Let \( O \) be the center of \( C_1 \) and \( C_2 \).
   - The radius of \( C_1 \) is \( \frac{1}{2} \) since its diameter is 1.
   - The radius of \( C_2 \) is \( \frac{k}{2} \) since its diameter is \( k \).
   - The radius of \( C_3 \) is \( k \) since its diameter is \( 2k \).

2. **Identify the points of intersection:**
   - The line \( AB \) intersects \( C_2 \) at points \( D \) and \( C \), with \( A \) between \( D \) and \( B \).
   - The line segment \( XY \) intersects \( C_2 \) again at point \( Z \).

3. **Calculate distances:**
   - \( OA = OB = \frac{1}{2} \)
   - \( OC = OD = \frac{k}{2} \)
   - \( DA = BC = \frac{k-1}{2} \)

4. **Determine the position of \( Z \):**
   - Since \( AY = k \) and \( OY = \frac{k}{2} \), \( Z \) is the midpoint of \( BY \).
   - Therefore, \( BZ = ZY \).

5. **Apply the power of a point theorem:**
   - The power of point \( B \) with respect to \( C_2 \) is given by:
     \[
     DB \cdot BC = XB \cdot BZ
     \]
   - Substituting the known distances:
     \[
     \left( \frac{k+1}{2} \right) \cdot \left( \frac{k-1}{2} \right) = XB \cdot \frac{BY}{2}
     \]

6. **Simplify the equation:**
   \[
   \frac{(k+1)(k-1)}{4} = \frac{XB \cdot BY}{2}
   \]
   \[
   \frac{k^2 - 1}{4} = \frac{XB \cdot BY}{2}
   \]
   \[
   \frac{k^2 - 1}{2} = XB \cdot BY
   \]

7. **Minimize the length of \( XY \):**
   - Using the AM-GM inequality:
     \[
     XB \cdot BY \leq \left( \frac{XB + BY}{2} \right)^2
     \]
   - Substituting the previous result:
     \[
     \frac{k^2 - 1}{2} \leq \left( \frac{XY}{2} \right)^2
     \]
   - Simplifying:
     \[
     k^2 - 1 \leq \frac{XY^2}{2}
     \]
     \[
     2(k^2 - 1) \leq XY^2
     \]
     \[
     XY \geq \sqrt{2(k^2 - 1)}
     \]

8. **Equality condition:**
   - Equality in the AM-GM inequality holds when \( XB = BY \).
   - Therefore, the ratio \( \frac{XB}{BY} = 1 \).

The final answer is \( \boxed{ \frac{XB}{BY} = 1 } \).