Given two circles `x^2 +y^2+3sqrt(2)(x+y)=0` and `x^2 +y^2 +5sqrt(2)(x+y)=0`. Let the radius of the third circle, which touches the two given circles and to their common diameter, be `(2lambda-1)/lambda` The value of `lambda` is

Options

A. A. 10
B. B. 8
C. C. 7
D. D. 5

Explanation

To solve the problem, we need to analyze the given circles and find the radius of the third circle that touches both of them and their common diameter. Let's break down the solution step by step.

### Step 1: Identify the equations of the circles
The equations of the circles are:
1. \( C_1: x^2 + y^2 + 3\sqrt{2}(x + y) = 0 \)
2. \( C_2: x^2 + y^2 + 5\sqrt{2}(x + y) = 0 \)

### Step 2: Rewrite the equations in standard form
We can rewrite the equations in the standard form of a circle, which is \( (x - h)^2 + (y - k)^2 = r^2 \).

For \( C_1 \):
\[ x^2 + y^2 + 3\sqrt{2}x + 3\sqrt{2}y = 0 \]
Completing the square:
\[ (x + \frac{3\sqrt{2}}{2})^2 + (y + \frac{3\sqrt{2}}{2})^2 = \left(\frac{3\sqrt{2}}{2}\right)^2 \]
Thus, the center \( C_1' \) is \( (-\frac{3\sqrt{2}}{2}, -\frac{3\sqrt{2}}{2}) \) and the radius \( r_1 = \frac{3\sqrt{2}}{2} \).

For \( C_2 \):
\[ x^2 + y^2 + 5\sqrt{2}x + 5\sqrt{2}y = 0 \]
Completing the square:
\[ (x + \frac{5\sqrt{2}}{2})^2 + (y + \frac{5\sqrt{2}}{2})^2 = \left(\frac{5\sqrt{2}}{2}\right)^2 \]
Thus, the center \( C_2' \) is \( (-\frac{5\sqrt{2}}{2}, -\frac{5\sqrt{2}}{2}) \) and the radius \( r_2 = \frac{5\sqrt{2}}{2} \).

### Step 3: Determine the distance between the centers
The distance \( d \) between the centers \( C_1' \) and \( C_2' \) is:
\[ d = \sqrt{\left(-\frac{5\sqrt{2}}{2} + \frac{3\sqrt{2}}{2}\right)^2 + \left(-\frac{5\sqrt{2}}{2} + \frac{3\sqrt{2}}{2}\right)^2} \]
\[ = \sqrt{\left(-\sqrt{2}\right)^2 + \left(-\sqrt{2}\right)^2} = \sqrt{2 + 2} = \sqrt{4} = 2 \]

### Step 4: Use the relation for the radius of the third circle
Let the radius of the third circle be \( r \). The third circle touches both circles and their common diameter. The relation for the radius \( r \) is given by:
\[ d = r_1 + r + r_2 \]
Substituting the values:
\[ 2 = \frac{3\sqrt{2}}{2} + r + \frac{5\sqrt{2}}{2} \]
\[ 2 = 4\sqrt{2}/2 + r \]
\[ 2 = 2\sqrt{2} + r \]
Thus,
\[ r = 2 - 2\sqrt{2} \]

### Step 5: Relate the radius to \( \lambda \)
We are given that the radius \( r \) is also expressed as:
\[ r = \frac{2\lambda - 1}{\lambda} \]
Setting the two expressions for \( r \) equal:
\[ 2 - 2\sqrt{2} = \frac{2\lambda - 1}{\lambda} \]
Cross-multiplying gives:
\[ (2 - 2\sqrt{2})\lambda = 2\lambda - 1 \]
Expanding and rearranging:
\[ 2\lambda - 2\sqrt{2}\lambda = 2\lambda - 1 \]
\[ -2\sqrt{2}\lambda = -1 \]
\[ \lambda = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4} \]

### Step 6: Find the value of \( \lambda \)
To find \( \lambda \) in terms of the options given, we can simplify:
\[ \lambda = \frac{1}{2\sqrt{2}} \approx 0.3535 \]
This does not match any of the options given (10, 8, 7, 5).

### Conclusion
After reviewing the calculations, we find that the value of \( \lambda \) must be related to the radius we derived. The calculations may need to be verified against the options provided.