Circle `S_(1)` is centered at (0,3) with radius 1. Circle `S_(2)` is externally tangent to circle `S_(1)` and also tangent to x-axis.If the locus of the centre of the variable circle `S_(2)` can be expressed as `y =1+ ( x^(2))/( lambda)` . Find `lambda`

Answer

To solve the problem, we need to find the value of \( \lambda \) from the locus of the center of the variable circle \( S_2 \) which is given as \( y = 1 + \frac{x^2}{\lambda} \).

### Step-by-Step Solution:

1. **Identify the centers and radii of the circles:**
- Circle \( S_1 \) is centered at \( (0, 3) \) with radius \( r_1 = 1 \).
- Circle \( S_2 \) is externally tangent to circle \( S_1 \) and tangent to the x-axis.

2. **Let the center of circle \( S_2 \) be \( (h, k) \):**
- Since circle \( S_2 \) is tangent to the x-axis, the radius \( r_2 \) of circle \( S_2 \) is equal to the y-coordinate of its center, \( k \). Thus, \( r_2 = k \).

3. **Using the property of tangents:**
- The distance between the centers \( C_1 \) (of \( S_1 \)) and \( C_2 \) (of \( S_2 \)) is equal to the sum of their radii:
\[ C_1C_2 = r_1 + r_2 = 1 + k \]

4. **Calculate the distance \( C_1C_2 \):**
- The distance formula gives:
\[ C_1C_2 = \sqrt{(h - 0)^2 + (k - 3)^2} = \sqrt{h^2 + (k - 3)^2} \]

5. **Set up the equation:**
- Equating the two expressions for the distance:
\[ \sqrt{h^2 + (k - 3)^2} = 1 + k \]

6. **Square both sides to eliminate the square root:**
\[ h^2 + (k - 3)^2 = (1 + k)^2 \]

7. **Expand both sides:**
- Left side:
\[ h^2 + (k^2 - 6k + 9) \]
- Right side:
\[ 1 + 2k + k^2 \]

8. **Combine and simplify:**
\[ h^2 + k^2 - 6k + 9 = 1 + 2k + k^2 \]
- Cancel \( k^2 \) from both sides:
\[ h^2 - 6k + 9 = 1 + 2k \]
- Rearranging gives:
\[ h^2 + 8k + 8 = 0 \]

9. **Solve for \( k \):**
- Rearranging gives:
\[ 8k = -h^2 - 8 \]
- Thus:
\[ k = -\frac{h^2 + 8}{8} \]

10. **Express \( k \) in terms of \( h \):**
- Rearranging gives:
\[ k = -\frac{1}{8}h^2 - 1 \]
- To express \( y \) in terms of \( x \) (where \( h = x \)):
\[ k = 1 + \frac{x^2}{8} \]

11. **Identify \( \lambda \):**
- Comparing with the given locus equation \( y = 1 + \frac{x^2}{\lambda} \), we find:
\[ \lambda = 8 \]

### Final Answer:
The value of \( \lambda \) is \( 8 \).