In the above problem, the complete range of the expression `x^(2)+y^(2)-26x+12y+210` is [a, b], then `b-2a=`

Answer

To solve the problem, we need to find the complete range of the expression \( x^2 + y^2 - 26x + 12y + 210 \) and then calculate \( b - 2a \) where the range is given as \([a, b]\).

### Step-by-Step Solution:

1. **Identify the expression**:
We start with the expression:
\[ f(x, y) = x^2 + y^2 - 26x + 12y + 210 \]

2. **Complete the square for \(x\)**:
To simplify the expression, we complete the square for \(x\):
\[ x^2 - 26x = (x - 13)^2 - 169 \]

3. **Complete the square for \(y\)**:
Next, we complete the square for \(y\):
\[ y^2 + 12y = (y + 6)^2 - 36 \]

4. **Rewrite the expression**:
Substituting the completed squares back into the expression, we have:
\[ f(x, y) = (x - 13)^2 - 169 + (y + 6)^2 - 36 + 210 \]
Simplifying this, we get:
\[ f(x, y) = (x - 13)^2 + (y + 6)^2 + 5 \]

5. **Determine the minimum value**:
The term \((x - 13)^2 + (y + 6)^2\) represents the distance squared from the point \((13, -6)\). This term is always non-negative and achieves its minimum value of \(0\) when \(x = 13\) and \(y = -6\). Thus, the minimum value of \(f(x, y)\) is:
\[ f(13, -6) = 0 + 5 = 5 \]

6. **Determine the range**:
Since \((x - 13)^2 + (y + 6)^2\) can take any non-negative value, the complete range of \(f(x, y)\) is:
\[ [5, \infty) \]
Therefore, we have \(a = 5\) and \(b = \infty\).

7. **Calculate \(b - 2a\)**:
Now we need to find \(b - 2a\):
\[ b - 2a = \infty - 2 \times 5 = \infty - 10 = \infty \]

### Final Answer:
Thus, the value of \(b - 2a\) is:
\[ \infty \]