lf the circle `x^(2)+y^(2)=2` and `x^(2)+y^(2)-4x-4y+lamda=0` have exactly three real common tangents then `lamda=`

Options

A. `-10`
B. 6
C. `-6`
D. 10

Answer

To find the value of \(\lambda\) such that the circles \(x^2 + y^2 = 2\) and \(x^2 + y^2 - 4x - 4y + \lambda = 0\) have exactly three real common tangents, we will follow these steps:

### Step 1: Identify the centers and radii of the circles

1. The first circle is given by the equation \(x^2 + y^2 = 2\).
- Center, \(C_1 = (0, 0)\)
- Radius, \(r_1 = \sqrt{2}\)

2. The second circle can be rewritten as:
\[ x^2 + y^2 - 4x - 4y + \lambda = 0 \implies (x^2 - 4x) + (y^2 - 4y) + \lambda = 0 \]
Completing the square:
\[ (x - 2)^2 - 4 + (y - 2)^2 - 4 + \lambda = 0 \implies (x - 2)^2 + (y - 2)^2 = 8 - \lambda \]
- Center, \(C_2 = (2, 2)\)
- Radius, \(r_2 = \sqrt{8 - \lambda}\)

### Step 2: Calculate the distance between the centers

The distance \(d\) between the centers \(C_1\) and \(C_2\) is calculated as:
\[ d = \sqrt{(2 - 0)^2 + (2 - 0)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]

### Step 3: Use the condition for three common tangents

For two circles to have exactly three common tangents, the distance between their centers must equal the sum of their radii:
\[ d = r_1 + r_2 \]
Substituting the known values:
\[ 2\sqrt{2} = \sqrt{2} + \sqrt{8 - \lambda} \]

### Step 4: Solve for \(\lambda\)

Rearranging the equation:
\[ 2\sqrt{2} - \sqrt{2} = \sqrt{8 - \lambda} \]
\[ \sqrt{2} = \sqrt{8 - \lambda} \]

Now squaring both sides:
\[ 2 = 8 - \lambda \]
Rearranging gives:
\[ \lambda = 8 - 2 = 6 \]

### Conclusion

Thus, the value of \(\lambda\) is:
\[ \lambda = 6 \]

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