Find all pairs $(a,b)$ of real numbers such that
$$|\sqrt{1-x^2 }-ax-b| \leq \frac{\sqrt{2} -1}{2}$$
holds for all $x\in [0,1]$.
1. **Substitute \( x = \cos t \) where \( t \in \left[ 0, \frac{\pi}{2} \right] \):**
   \[
   \sqrt{1 - x^2} = \sqrt{1 - \cos^2 t} = \sin t
   \]
   The given inequality becomes:
   \[
   |\sin t - a \cos t - b| \leq \frac{\sqrt{2} - 1}{2}
   \]

2. **Express the trigonometric terms in a single cosine function:**
   \[
   \sin t - a \cos t = \sqrt{1 + a^2} \left( \frac{1}{\sqrt{1 + a^2}} \sin t - \frac{a}{\sqrt{1 + a^2}} \cos t \right)
   \]
   Let:
   \[
   \cos \alpha = \frac{1}{\sqrt{1 + a^2}}, \quad \sin \alpha = \frac{a}{\sqrt{1 + a^2}}
   \]
   Then:
   \[
   \sin t - a \cos t = \sqrt{1 + a^2} \left( \cos \alpha \sin t - \sin \alpha \cos t \right) = \sqrt{1 + a^2} \sin (t - \alpha)
   \]

3. **Rewrite the inequality:**
   \[
   |\sqrt{1 + a^2} \sin (t - \alpha) - b| \leq \frac{\sqrt{2} - 1}{2}
   \]

4. **Determine the maximum and minimum values of \(\sin (t - \alpha)\):**
   Since \(\sin (t - \alpha)\) ranges from \(-1\) to \(1\), the expression \(\sqrt{1 + a^2} \sin (t - \alpha)\) ranges from \(-\sqrt{1 + a^2}\) to \(\sqrt{1 + a^2}\).

5. **Analyze the bounds:**
   \[
   \left| \sqrt{1 + a^2} - b \right| \leq \frac{\sqrt{2} - 1}{2}
   \]
   \[
   \left| -\sqrt{1 + a^2} - b \right| \leq \frac{\sqrt{2} - 1}{2}
   \]

6. **Combine the inequalities:**
   \[
   -\frac{\sqrt{2} - 1}{2} \leq \sqrt{1 + a^2} - b \leq \frac{\sqrt{2} - 1}{2}
   \]
   \[
   -\frac{\sqrt{2} - 1}{2} \leq -\sqrt{1 + a^2} - b \leq \frac{\sqrt{2} - 1}{2}
   \]

7. **Solve the inequalities:**
   \[
   -\frac{\sqrt{2} - 1}{2} \leq \sqrt{1 + a^2} - b \leq \frac{\sqrt{2} - 1}{2}
   \]
   \[
   -\frac{\sqrt{2} - 1}{2} \leq -\sqrt{1 + a^2} - b \leq \frac{\sqrt{2} - 1}{2}
   \]

   From the first inequality:
   \[
   b - \frac{\sqrt{2} - 1}{2} \leq \sqrt{1 + a^2} \leq b + \frac{\sqrt{2} - 1}{2}
   \]

   From the second inequality:
   \[
   b + \frac{\sqrt{2} - 1}{2} \geq -\sqrt{1 + a^2} \geq b - \frac{\sqrt{2} - 1}{2}
   \]

8. **Combine the results:**
   \[
   b - \frac{\sqrt{2} - 1}{2} \leq \sqrt{1 + a^2} \leq b + \frac{\sqrt{2} - 1}{2}
   \]
   \[
   b + \frac{\sqrt{2} - 1}{2} \geq -\sqrt{1 + a^2} \geq b - \frac{\sqrt{2} - 1}{2}
   \]

   These inequalities imply that \( b \) must be zero for the inequalities to hold for all \( a \).

9. **Conclusion:**
   \[
   b = 0 \quad \text{and} \quad \sqrt{1 + a^2} \leq \frac{\sqrt{2} - 1}{2}
   \]

   Solving for \( a \):
   \[
   1 + a^2 \leq \left( \frac{\sqrt{2} - 1}{2} \right)^2
   \]
   \[
   1 + a^2 \leq \frac{3 - 2\sqrt{2}}{4}
   \]
   \[
   a^2 \leq \frac{3 - 2\sqrt{2}}{4} - 1
   \]
   \[
   a^2 \leq \frac{-1 - 2\sqrt{2}}{4}
   \]

   Since \( a^2 \) cannot be negative, there are no real solutions for \( a \).

The final answer is \( \boxed{ (a, b) = (0, 0) } \).