If the radius of the circle passing through the origin and touching the line `x+y=2` at `(1, 1)` is r units, then the value of `3sqrt2r` is

Answer

To solve the problem, we need to find the radius \( r \) of a circle that passes through the origin and touches the line \( x + y = 2 \) at the point \( (1, 1) \). Once we find \( r \), we will calculate \( 3\sqrt{2}r \).

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The circle passes through the origin \( (0, 0) \).
- The circle touches the line \( x + y = 2 \) at the point \( (1, 1) \).

2. **Find the Slope of the Line:**
- The equation of the line can be rewritten as \( y = -x + 2 \).
- The slope \( m_1 \) of this line is \( -1 \).

3. **Find the Slope of the Line from the Origin to the Point of Tangency:**
- The slope \( m_2 \) of the line connecting the origin \( (0, 0) \) and the point \( (1, 1) \) is calculated as:
\[ m_2 = \frac{1 - 0}{1 - 0} = 1 \]

4. **Determine the Relationship Between the Slopes:**
- Since the line \( x + y = 2 \) is tangent to the circle at \( (1, 1) \), and the line from the origin to \( (1, 1) \) is perpendicular to the tangent line, we have:
\[ m_1 \times m_2 = -1 \quad \text{(the product of slopes of perpendicular lines)} \]
- This confirms that \( -1 \times 1 = -1 \).

5. **Find the Length of the Diameter:**
- The distance \( AB \) (where \( A \) is the origin \( (0, 0) \) and \( B \) is the point of tangency \( (1, 1) \)) can be calculated using the distance formula:
\[ AB = \sqrt{(1 - 0)^2 + (1 - 0)^2} = \sqrt{1 + 1} = \sqrt{2} \]
- Since \( AB \) is the diameter of the circle, we have:
\[ D = \sqrt{2} \]

6. **Calculate the Radius:**
- The radius \( r \) is half of the diameter:
\[ r = \frac{D}{2} = \frac{\sqrt{2}}{2} \]

7. **Calculate \( 3\sqrt{2}r \):**
- Now we substitute \( r \) into the expression \( 3\sqrt{2}r \):
\[ 3\sqrt{2}r = 3\sqrt{2} \cdot \frac{\sqrt{2}}{2} = 3 \cdot \frac{2}{2} = 3 \]

### Final Answer:
The value of \( 3\sqrt{2}r \) is \( 3 \).