The circle `x^2+y^2+4lamdax=0` which `lamda in R` touches the parabola `y^2=8x`. The value of `lamda` is given by

Answer

To solve the problem, we need to find the value of \(\lambda\) such that the circle given by the equation \(x^2 + y^2 + 4\lambda x = 0\) touches the parabola given by the equation \(y^2 = 8x\).

### Step-by-Step Solution:

1. **Write down the equations**:
- Circle: \(x^2 + y^2 + 4\lambda x = 0\)
- Parabola: \(y^2 = 8x\)

2. **Substitute the parabola's equation into the circle's equation**:
Since the parabola touches the circle, we can substitute \(y^2\) from the parabola into the circle's equation:
\[ x^2 + 8x + 4\lambda x = 0 \]
This simplifies to:
\[ x^2 + (8 + 4\lambda)x = 0 \]

3. **Factor the equation**:
We can factor out \(x\):
\[ x(x + (8 + 4\lambda)) = 0 \]
This gives us two solutions: \(x = 0\) or \(x + (8 + 4\lambda) = 0\).

4. **Condition for tangency**:
For the circle to touch the parabola, the quadratic equation must have exactly one solution. This occurs when the discriminant \(D\) is equal to zero. The discriminant for the quadratic \(x^2 + (8 + 4\lambda)x = 0\) is:
\[ D = b^2 - 4ac \]
Here, \(a = 1\), \(b = 8 + 4\lambda\), and \(c = 0\). Thus, the discriminant is:
\[ D = (8 + 4\lambda)^2 - 4 \cdot 1 \cdot 0 = (8 + 4\lambda)^2 \]

5. **Set the discriminant to zero**:
For tangency, we set the discriminant equal to zero:
\[ (8 + 4\lambda)^2 = 0 \]

6. **Solve for \(\lambda\)**:
Taking the square root of both sides gives:
\[ 8 + 4\lambda = 0 \]
Solving for \(\lambda\):
\[ 4\lambda = -8 \implies \lambda = -2 \]

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

---