If a circle S(x,y)=0 touches the point (2,3) of the line x+y=5 and S(1,2)=0, then radius of such circle is `(1)/(sqrtlamda)` units then the value of `lamda` is.

Answer

To solve the problem, we need to find the value of \( \lambda \) given that the radius of the circle is \( \frac{1}{\sqrt{\lambda}} \) units. We know that the circle touches the point (2,3) on the line \( x + y = 5 \) and passes through the point (1,2).

### Step-by-Step Solution:

1. **Identify the Given Information**:
- The line equation is \( x + y = 5 \).
- The circle touches the point \( (2, 3) \) on this line.
- The circle passes through the point \( (1, 2) \).

2. **Find the Slope of the Line**:
- Rewriting the line equation in slope-intercept form:
\[ y = -x + 5 \]
- The slope of the line is \( -1 \).

3. **Determine the Slope of the Perpendicular Line**:
- The slope of the line perpendicular to the given line is the negative reciprocal of \( -1 \), which is \( 1 \).

4. **Equation of the Perpendicular Line**:
- The perpendicular line passes through the point \( (2, 3) \) and has a slope of \( 1 \).
- Using the point-slope form of the line equation:
\[ y - 3 = 1(x - 2) \]
- Simplifying this gives:
\[ y - 3 = x - 2 \implies x - y = -1 \implies x - y = 1 \]

5. **Find the Midpoint of the Diameter**:
- Since \( (1, 2) \) and \( (2, 3) \) are diametrically opposite points on the circle, we can find the center of the circle by calculating the midpoint of these two points.
- The midpoint \( M \) is given by:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{1 + 2}{2}, \frac{2 + 3}{2} \right) = \left( \frac{3}{2}, \frac{5}{2} \right) \]

6. **Calculate the Radius of the Circle**:
- The radius \( r \) is half the distance between the two points \( (1, 2) \) and \( (2, 3) \).
- Using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(2 - 1)^2 + (3 - 2)^2} = \sqrt{1 + 1} = \sqrt{2} \]
- Therefore, the radius \( r \) is:
\[ r = \frac{d}{2} = \frac{\sqrt{2}}{2} \]

7. **Express the Radius in the Given Form**:
- We know that the radius can also be expressed as \( \frac{1}{\sqrt{\lambda}} \).
- Setting the two expressions for the radius equal gives:
\[ \frac{1}{\sqrt{\lambda}} = \frac{\sqrt{2}}{2} \]
- Cross-multiplying yields:
\[ 2 = \sqrt{2 \lambda} \]
- Squaring both sides results in:
\[ 4 = 2\lambda \implies \lambda = 2 \]

### Final Answer:
The value of \( \lambda \) is \( 2 \).