In a coordinate system, a circle with radius $7$ and center is on the y-axis placed inside the parabola with equation $y = x^2$ , so that it just touches the parabola in two points. Determine the coordinate set for the center of the circle.

1. **Symmetry and Circle Equation**:
   By symmetry, the center of the circle must lie on the y-axis, i.e., $x = 0$. Therefore, the equation of the circle can be written as:
   \[
   x^2 + (y - n)^2 = 49
   \]
   where $n$ is the y-coordinate of the center of the circle.

2. **Intersection Points**:
   The circle touches the parabola $y = x^2$ at two points, which by symmetry are $(a, b)$ and $(-a, b)$. Substituting $y = x^2$ into the circle's equation, we get:
   \[
   x^2 + (x^2 - n)^2 = 49
   \]
   Simplifying, we have:
   \[
   x^2 + (x^4 - 2nx^2 + n^2) = 49
   \]
   \[
   x^4 - x^2(2n - 1) + n^2 - 49 = 0
   \]

3. **Quadratic in $y$**:
   Let $y = x^2$. Then the equation becomes:
   \[
   y + (y - n)^2 = 49
   \]
   \[
   y + y^2 - 2ny + n^2 = 49
   \]
   \[
   y^2 + (1 - 2n)y + n^2 - 49 = 0
   \]

4. **Discriminant Condition**:
   Since the circle touches the parabola at exactly two points with the same y-coordinate, the quadratic equation in $y$ must have exactly one solution. Therefore, the discriminant of the quadratic equation must be zero:
   \[
   \Delta = (1 - 2n)^2 - 4(n^2 - 49) = 0
   \]
   Simplifying the discriminant:
   \[
   (1 - 2n)^2 - 4(n^2 - 49) = 0
   \]
   \[
   1 - 4n + 4n^2 - 4n^2 + 196 = 0
   \]
   \[
   1 - 4n + 196 = 0
   \]
   \[
   197 - 4n = 0
   \]
   \[
   4n = 197
   \]
   \[
   n = \frac{197}{4}
   \]

5. **Conclusion**:
   Thus, the center of the circle is at:
   \[
   (0, \frac{197}{4})
   \]

The final answer is \( \boxed{ (0, \frac{197}{4}) } \)