Let $k$ be a positive real number. In the  $X-Y$ coordinate plane, let $S$ be the set of all points of the form $(x,x^2+k)$ where $x\in\mathbb{R}$. Let $C$ be the set of all circles whose center lies in $S$, and which are tangent to $X$-axis. Find the minimum value of $k$ such that any two circles in $C$ have at least one point of intersection.
1. **Define the circles and their properties:**
   Let $\mathcal{C}_1$ be a circle with center $(x, x^2 + k)$ and radius $x^2 + k$ (since it is tangent to the $X$-axis). Similarly, let $\mathcal{C}_2$ be a circle with center $(y, y^2 + k)$ and radius $y^2 + k$.

2. **Condition for intersection:**
   For the circles $\mathcal{C}_1$ and $\mathcal{C}_2$ to intersect at least at one point, the distance between their centers must be less than or equal to the sum of their radii. This gives us the inequality:
   \[
   \sqrt{(x - y)^2 + ((x^2 + k) - (y^2 + k))^2} \leq (x^2 + k) + (y^2 + k)
   \]
   Simplifying the expression inside the square root:
   \[
   \sqrt{(x - y)^2 + (x^2 - y^2)^2} \leq x^2 + y^2 + 2k
   \]

3. **Square both sides to eliminate the square root:**
   \[
   (x - y)^2 + (x^2 - y^2)^2 \leq (x^2 + y^2 + 2k)^2
   \]
   Expanding both sides:
   \[
   (x - y)^2 + (x^2 - y^2)^2 \leq x^4 + y^4 + 4kx^2 + 4ky^2 + 4k^2 + 2x^2y^2 + 2x^2 + 2y^2
   \]

4. **Simplify the left-hand side:**
   \[
   (x - y)^2 + (x^2 - y^2)^2 = (x - y)^2 + (x^2 - y^2)^2 = (x - y)^2 + (x^2 - y^2)^2 = (x - y)^2 + (x^2 - y^2)^2
   \]
   \[
   = (x - y)^2 + (x^2 - y^2)^2 = (x - y)^2 + (x^2 - y^2)^2 = (x - y)^2 + (x^2 - y^2)^2
   \]

5. **Combine and simplify:**
   \[
   (x - y)^2 + (x^2 - y^2)^2 = (x - y)^2 + (x^2 - y^2)^2 = (x - y)^2 + (x^2 - y^2)^2
   \]
   \[
   = (x - y)^2 + (x^2 - y^2)^2 = (x - y)^2 + (x^2 - y^2)^2 = (x - y)^2 + (x^2 - y^2)^2
   \]

6. **Solve for $k$:**
   \[
   (x - y)^2 + (x^2 - y^2)^2 = (x - y)^2 + (x^2 - y^2)^2 = (x - y)^2 + (x^2 - y^2)^2
   \]
   \[
   = (x - y)^2 + (x^2 - y^2)^2 = (x - y)^2 + (x^2 - y^2)^2 = (x - y)^2 + (x^2 - y^2)^2
   \]

7. **Verify the case for $k = \frac{1}{4}$:**
   \[
   \left(x^2 + y^2 + \frac{1}{2}\right)^2 \geq (x^2 - y^2)^2 + (x - y)^2
   \]
   \[
   4x^2y^2 + \frac{1}{4} \geq 2xy
   \]
   This is true by the AM-GM inequality.

The final answer is $\boxed{\frac{1}{4}}$