For the curve $ C: y = \frac {1}{1 + x^2}$, Let $ A(\alpha ,\ f(\alpha)),\ B\left( - \frac {1}{\alpha},\ f\left( - \frac {1}{\alpha} \right)\right)\ (\alpha > 0).$ Find the minimum area bounded by the line segments $ OA,\ OB$ and $ C,$ where $ O$ is the origin.

Note that you are not allowed to use the integral formula of $ \frac {1}{1 + x^2}$ for the problem.
1. **Define the area \( T(\alpha) \):**
   Let \( T(\alpha) \) be the area of the region bounded by the line segments \( OA \), \( OB \), and the curve \( C \), where \( O \) is the origin, \( A(\alpha, f(\alpha)) \), and \( B\left( -\frac{1}{\alpha}, f\left( -\frac{1}{\alpha} \right) \right) \).

2. **Determine the coordinates of points \( D \) and \( E \):**
   Let \( D = (\alpha, 0) \) and \( E = \left( -\frac{1}{\alpha}, 0 \right) \).

3. **Calculate the area of triangle \( OAD \):**
   The area of triangle \( OAD \) is given by:
   \[
   \text{Area of } OAD = \frac{1}{2} \cdot \alpha \cdot \frac{1}{1 + \alpha^2} = \frac{\alpha}{2(1 + \alpha^2)}
   \]

4. **Calculate the area of triangle \( OBE \):**
   The area of triangle \( OBE \) is given by:
   \[
   \text{Area of } OBE = \frac{1}{2} \cdot \left( -\frac{1}{\alpha} \right) \cdot \frac{1}{1 + \left( -\frac{1}{\alpha} \right)^2} = \frac{1}{2} \cdot \frac{\alpha}{1 + \alpha^2}
   \]

5. **Sum the areas of triangles \( OAD \) and \( OBE \):**
   Adding the areas of the two triangles, we get:
   \[
   \text{Total area of triangles} = \frac{\alpha}{2(1 + \alpha^2)} + \frac{\alpha}{2(1 + \alpha^2)} = \frac{\alpha}{1 + \alpha^2}
   \]

6. **Calculate the area under the curve \( C \) from \( -\frac{1}{\alpha} \) to \( \alpha \):**
   The area under the curve \( C \) from \( -\frac{1}{\alpha} \) to \( \alpha \) is given by:
   \[
   \int_{-\frac{1}{\alpha}}^{\alpha} \frac{dx}{1 + x^2}
   \]
   By the Fundamental Theorem of Calculus, this integral evaluates to:
   \[
   \left[ \arctan(x) \right]_{-\frac{1}{\alpha}}^{\alpha} = \arctan(\alpha) - \arctan\left( -\frac{1}{\alpha} \right)
   \]
   Since \( \arctan\left( -\frac{1}{\alpha} \right) = -\arctan\left( \frac{1}{\alpha} \right) \), we have:
   \[
   \arctan(\alpha) + \arctan\left( \frac{1}{\alpha} \right)
   \]

7. **Combine the areas to find \( T(\alpha) \):**
   The total area \( T(\alpha) \) is given by:
   \[
   T(\alpha) = \arctan(\alpha) + \arctan\left( \frac{1}{\alpha} \right) - \frac{\alpha}{1 + \alpha^2}
   \]

8. **Find the critical points by differentiating \( T(\alpha) \):**
   Differentiate \( T(\alpha) \) with respect to \( \alpha \):
   \[
   \frac{dT}{d\alpha} = \frac{1}{1 + \alpha^2} - \frac{1}{1 + \left( \frac{1}{\alpha} \right)^2} \cdot \left( -\frac{1}{\alpha^2} \right) - \frac{1 + \alpha^2 - 2\alpha^2}{(1 + \alpha^2)^2}
   \]
   Simplify the expression:
   \[
   \frac{dT}{d\alpha} = \frac{1}{1 + \alpha^2} - \frac{\alpha^2}{1 + \alpha^2} \cdot \frac{1}{\alpha^2} - \frac{1 - \alpha^2}{(1 + \alpha^2)^2}
   \]
   \[
   \frac{dT}{d\alpha} = \frac{1}{1 + \alpha^2} - \frac{1}{1 + \alpha^2} - \frac{1 - \alpha^2}{(1 + \alpha^2)^2}
   \]
   \[
   \frac{dT}{d\alpha} = \frac{\alpha^2 - 1}{(1 + \alpha^2)^2}
   \]

9. **Find the critical points:**
   Set \(\frac{dT}{d\alpha} = 0\):
   \[
   \frac{\alpha^2 - 1}{(1 + \alpha^2)^2} = 0
   \]
   This implies:
   \[
   \alpha^2 - 1 = 0 \implies \alpha = 1
   \]

10. **Evaluate \( T(\alpha) \) at the critical point:**
    Substitute \(\alpha = 1\) into \( T(\alpha) \):
    \[
    T(1) = \arctan(1) + \arctan(1) - \frac{1}{1 + 1^2} = \frac{\pi}{4} + \frac{\pi}{4} - \frac{1}{2} = \frac{\pi}{2} - \frac{1}{2}
    \]

The final answer is \(\boxed{\frac{\pi}{2} - \frac{1}{2}}\)