If the equation `3x^2 + 3y^2 + 6lambdax + 2lambda = 0` represents a circle, then the value of `lambda`. lies in

Options

A. `(oo, 0) uu[(2)/(3),oo]`
B. `(-oo,-1)uu [(1)/(3), oo)`
C. `(-∞, 0]∪ [(2)/(3) ,∞)
D. none of these

Answer

To determine the values of \( \lambda \) for which the equation \( 3x^2 + 3y^2 + 6\lambda x + 2\lambda = 0 \) represents a circle, we can follow these steps:

### Step 1: Simplify the Equation
We start by dividing the entire equation by 3 to simplify it:
\[ x^2 + y^2 + 2\lambda x + \frac{2\lambda}{3} = 0 \]

### Step 2: Rearrange the Equation
Next, we rearrange the equation to isolate the constant term:
\[ x^2 + y^2 + 2\lambda x = -\frac{2\lambda}{3} \]

### Step 3: Complete the Square
To express the equation in the standard form of a circle, we complete the square for the \( x \) terms:
\[ (x^2 + 2\lambda x) + y^2 = -\frac{2\lambda}{3} \]
Completing the square for \( x \):
\[ (x + \lambda)^2 - \lambda^2 + y^2 = -\frac{2\lambda}{3} \]
This leads to:
\[ (x + \lambda)^2 + y^2 = \lambda^2 - \frac{2\lambda}{3} \]

### Step 4: Identify the Condition for a Circle
For the equation to represent a circle, the right-hand side must be non-negative (since it represents \( r^2 \)):
\[ \lambda^2 - \frac{2\lambda}{3} \geq 0 \]

### Step 5: Factor the Inequality
We can factor the left-hand side:
\[ \lambda \left( \lambda - \frac{2}{3} \right) \geq 0 \]

### Step 6: Determine the Critical Points
The critical points from the inequality are \( \lambda = 0 \) and \( \lambda = \frac{2}{3} \).

### Step 7: Test Intervals
We analyze the sign of the expression \( \lambda \left( \lambda - \frac{2}{3} \right) \) in the intervals:
1. \( (-\infty, 0) \)
2. \( (0, \frac{2}{3}) \)
3. \( (\frac{2}{3}, \infty) \)

- For \( \lambda < 0 \): Both factors are negative, so the product is positive.
- For \( 0 < \lambda < \frac{2}{3} \): The first factor is positive and the second is negative, so the product is negative.
- For \( \lambda > \frac{2}{3} \): Both factors are positive, so the product is positive.

### Step 8: Conclusion
The solution to the inequality \( \lambda \left( \lambda - \frac{2}{3} \right) \geq 0 \) is:
\[ \lambda \in (-\infty, 0] \cup \left[\frac{2}{3}, \infty\right) \]

### Final Answer
Thus, the values of \( \lambda \) for which the equation represents a circle are:
\[ \lambda \in (-\infty, 0] \cup \left[\frac{2}{3}, \infty\right) \]

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