Let $ n \in \mathbb{Z}^\plus{}$ and let $ a, b \in \mathbb{R}.$ Determine the range of $ x_0$ for which

\[ \sum^n_{i\equal{}0} x_i \equal{} a \text{ and } \sum^n_{i\equal{}0} x^2_i \equal{} b,\]

where $ x_0, x_1, \ldots , x_n$ are real variables.
1. **Apply the Power-Mean Inequality:**
   The Power-Mean Inequality states that for any non-negative real numbers \( x_1, x_2, \ldots, x_n \) and any real number \( p \geq q \),
   \[
   \left( \frac{1}{n} \sum_{i=1}^n x_i^p \right)^{\frac{1}{p}} \geq \left( \frac{1}{n} \sum_{i=1}^n x_i^q \right)^{\frac{1}{q}}.
   \]
   In our case, we use \( p = 2 \) and \( q = 1 \):
   \[
   \left( \frac{1}{n} \sum_{i=1}^n x_i^2 \right)^{\frac{1}{2}} \geq \left( \frac{1}{n} \sum_{i=1}^n x_i \right).
   \]
   Squaring both sides, we get:
   \[
   \frac{1}{n} \sum_{i=1}^n x_i^2 \geq \left( \frac{1}{n} \sum_{i=1}^n x_i \right)^2.
   \]

2. **Express the sums in terms of \( x_0 \):**
   Given the conditions:
   \[
   \sum_{i=0}^n x_i = a \quad \text{and} \quad \sum_{i=0}^n x_i^2 = b,
   \]
   we can separate \( x_0 \) from the sums:
   \[
   \sum_{i=1}^n x_i = a - x_0 \quad \text{and} \quad \sum_{i=1}^n x_i^2 = b - x_0^2.
   \]

3. **Apply the Power-Mean Inequality to the remaining terms:**
   Substitute these into the Power-Mean Inequality:
   \[
   \frac{b - x_0^2}{n} \geq \left( \frac{a - x_0}{n} \right)^2.
   \]
   Simplify the inequality:
   \[
   b - x_0^2 \geq \frac{(a - x_0)^2}{n}.
   \]
   Multiply both sides by \( n \):
   \[
   n(b - x_0^2) \geq (a - x_0)^2.
   \]

4. **Expand and rearrange the inequality:**
   Expand the right-hand side:
   \[
   n b - n x_0^2 \geq a^2 - 2 a x_0 + x_0^2.
   \]
   Combine like terms:
   \[
   (n + 1) x_0^2 - 2 a x_0 + (a^2 - n b) \leq 0.
   \]

5. **Solve the quadratic inequality:**
   This is a quadratic inequality in \( x_0 \):
   \[
   (n + 1) x_0^2 - 2 a x_0 + (a^2 - n b) \leq 0.
   \]
   The roots of the corresponding quadratic equation are:
   \[
   x_0 = \frac{2a \pm \sqrt{4a^2 - 4(n+1)(a^2 - n b)}}{2(n+1)} = \frac{a \pm \sqrt{n((n+1)b - a^2)}}{n+1}.
   \]
   Let \( D = \sqrt{n((n+1)b - a^2)} \). The roots are:
   \[
   x_0 = \frac{a \pm D}{n+1}.
   \]

6. **Determine the range of \( x_0 \):**
   The quadratic inequality \( (n + 1) x_0^2 - 2 a x_0 + (a^2 - n b) \leq 0 \) holds for:
   \[
   \frac{a - D}{n+1} \leq x_0 \leq \frac{a + D}{n+1}.
   \]

The final answer is \( \boxed{ \left[\frac{a - D}{n+1}, \frac{a + D}{n+1}\right] } \).