Let `A= {(x,y) in R xx R |2x^(2)+2y^(2)-2x-2y=1}, B= {(x,y) in R xx R|4x^(2)+4y^(2)-16y+7=0} and C={(x,y) in R xx R |x^(2)+y^(2)-4x-2y+5 le r^(2)}`. Then the minimum value of `|r|` such that `A uu B sube C` is equal to

Options

A. `(3+sqrt10)/(2)`
B. `(2+sqrt10)/(2)`
C. `(3+2sqrt5)/(2)`
D. `1+sqrt5`

Answer

To solve the problem, we need to analyze the sets \( A \), \( B \), and \( C \) given in the question.

### Step 1: Analyze Set A
The set \( A \) is defined by the equation:
\[ 2x^2 + 2y^2 - 2x - 2y = 1 \]
Dividing the entire equation by 2, we get:
\[ x^2 + y^2 - x - y = \frac{1}{2} \]
Rearranging this, we can complete the square:
\[ (x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = \frac{1}{2} + \frac{1}{4} + \frac{1}{4} = 1 \]
Thus, set \( A \) represents a circle centered at \( \left(\frac{1}{2}, \frac{1}{2}\right) \) with a radius of \( 1 \).

### Step 2: Analyze Set B
The set \( B \) is defined by the equation:
\[ 4x^2 + 4y^2 - 16y + 7 = 0 \]
Rearranging gives:
\[ 4x^2 + 4y^2 - 16y = -7 \]
Dividing by 4:
\[ x^2 + y^2 - 4y = -\frac{7}{4} \]
Completing the square for \( y \):
\[ x^2 + (y - 2)^2 = 4 - \frac{7}{4} = \frac{16}{4} - \frac{7}{4} = \frac{9}{4} \]
Thus, set \( B \) represents a circle centered at \( (0, 2) \) with a radius of \( \frac{3}{2} \).

### Step 3: Analyze Set C
The set \( C \) is defined by the inequality:
\[ x^2 + y^2 - 4x - 2y + 5 \leq r^2 \]
Rearranging gives:
\[ (x^2 - 4x) + (y^2 - 2y) \leq r^2 - 5 \]
Completing the square:
\[ (x - 2)^2 + (y - 1)^2 \leq r^2 - 5 + 4 + 1 \]
This simplifies to:
\[ (x - 2)^2 + (y - 1)^2 \leq r^2 \]
Thus, set \( C \) represents a circle centered at \( (2, 1) \) with radius \( r \).

### Step 4: Determine Minimum Value of \( |r| \)
To ensure that \( A \cup B \subseteq C \), we need to find the minimum value of \( r \) such that both circles \( A \) and \( B \) lie completely within circle \( C \).

1. **Distance from center of \( A \) to center of \( C \)**:
\[ d(A, C) = \sqrt{(2 - \frac{1}{2})^2 + (1 - \frac{1}{2})^2} = \sqrt{\left(\frac{3}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{9}{4} + \frac{1}{4}} = \sqrt{\frac{10}{4}} = \frac{\sqrt{10}}{2} \]
The radius of circle \( A \) is \( 1 \). Therefore, for \( A \) to be inside \( C \):
\[ r \geq d(A, C) + 1 = \frac{\sqrt{10}}{2} + 1 \]

2. **Distance from center of \( B \) to center of \( C \)**:
\[ d(B, C) = \sqrt{(2 - 0)^2 + (1 - 2)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \]
The radius of circle \( B \) is \( \frac{3}{2} \). Therefore, for \( B \) to be inside \( C \):
\[ r \geq d(B, C) + \frac{3}{2} = \sqrt{5} + \frac{3}{2} \]

### Step 5: Compare the Values
We need to find the maximum of the two conditions:
1. \( r \geq \frac{\sqrt{10}}{2} + 1 \)
2. \( r \geq \sqrt{5} + \frac{3}{2} \)

Calculating these values:
- \( \frac{\sqrt{10}}{2} + 1 \approx 1.58 + 1 = 2.58 \)
- \( \sqrt{5} + \frac{3}{2} \approx 2.24 + 1.5 = 3.74 \)

Thus, the minimum value of \( |r| \) such that \( A \cup B \subseteq C \) is:
\[ \boxed{3.74} \]