If a circle having centre at `(alpha,beta)` radius r completely lies with in two lines x+y=2 and x+y=-2, then, min. `(|alpha+beta+2|,|alpha+beta-2|` is

Options

A. greater than `sqrt2r`
B. less than `sqrt2r`
C. greater than 2r
D. less than 2r

Answer

To solve the problem, we need to find the minimum value of \( \min(|\alpha + \beta + 2|, |\alpha + \beta - 2|) \) given that the circle with center \((\alpha, \beta)\) and radius \(r\) lies completely within the lines \(x + y = 2\) and \(x + y = -2\).

### Step 1: Understand the Position of the Circle
The lines \(x + y = 2\) and \(x + y = -2\) are parallel lines. The distance between these two lines can be calculated as follows:
- The distance between two parallel lines \(Ax + By + C_1 = 0\) and \(Ax + By + C_2 = 0\) is given by the formula:
\[ \text{Distance} = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} \]
For our lines:
- \(C_1 = -2\) (from \(x + y - 2 = 0\))
- \(C_2 = 2\) (from \(x + y + 2 = 0\))
- Here, \(A = 1\), \(B = 1\)

Calculating the distance:
\[ \text{Distance} = \frac{|2 - (-2)|}{\sqrt{1^2 + 1^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \]

### Step 2: Determine the Constraints for the Circle
Since the circle must lie completely within these two lines, the distance from the center of the circle \((\alpha, \beta)\) to either line must be greater than or equal to the radius \(r\).

1. **Distance from the line \(x + y = 2\)**:
\[ d_1 = \frac{|\alpha + \beta - 2|}{\sqrt{2}} \geq r \]
This gives us:
\[ |\alpha + \beta - 2| \geq r\sqrt{2} \tag{1} \]

2. **Distance from the line \(x + y = -2\)**:
\[ d_2 = \frac{|\alpha + \beta + 2|}{\sqrt{2}} \geq r \]
This gives us:
\[ |\alpha + \beta + 2| \geq r\sqrt{2} \tag{2} \]

### Step 3: Combine the Inequalities
From inequalities (1) and (2), we have:
\[ |\alpha + \beta - 2| \geq r\sqrt{2} \]
\[ |\alpha + \beta + 2| \geq r\sqrt{2} \]

### Step 4: Find the Minimum Value
We need to find the minimum of \( \min(|\alpha + \beta + 2|, |\alpha + \beta - 2|) \). Since both expressions are greater than or equal to \(r\sqrt{2}\), we can conclude that:
\[ \min(|\alpha + \beta + 2|, |\alpha + \beta - 2|) \geq r\sqrt{2} \]

### Conclusion
The minimum value of \( \min(|\alpha + \beta + 2|, |\alpha + \beta - 2|) \) is \( r\sqrt{2} \).