If `3x^(2)+2lambda xy + 3y^(2)+(6-lambda)x+(2lambda-6)y-21=0` is the equation of a circle, then its radius is

Options

A. 1
B. 3
C. `2sqrt(2)`
D. none of these

Explanation

To determine the radius of the circle represented by the equation

\[ 3x^2 + 2\lambda xy + 3y^2 + (6 - \lambda)x + (2\lambda - 6)y - 21 = 0, \]

we need to ensure that this equation satisfies the conditions for a circle.

### Step-by-Step Solution:

1. **Identify Conditions for a Circle**:
For a general conic section to represent a circle, the following conditions must be met:
- The coefficient of \(x^2\) must be equal to the coefficient of \(y^2\).
- The coefficient of \(xy\) must be zero.

2. **Set Up the Conditions**:
From the given equation, we have:
- Coefficient of \(x^2\) = 3
- Coefficient of \(y^2\) = 3
- Coefficient of \(xy\) = \(2\lambda\)

For the equation to represent a circle:
\[ 3 = 3 \quad \text{(this is satisfied)} \]
\[ 2\lambda = 0 \quad \Rightarrow \quad \lambda = 0 \]

3. **Substitute \(\lambda\) into the Equation**:
Substitute \(\lambda = 0\) into the original equation:
\[ 3x^2 + 0 \cdot xy + 3y^2 + (6 - 0)x + (2 \cdot 0 - 6)y - 21 = 0 \]
Simplifying this gives:
\[ 3x^2 + 3y^2 + 6x - 6y - 21 = 0 \]

4. **Divide the Entire Equation by 3**:
To simplify, divide the entire equation by 3:
\[ x^2 + y^2 + 2x - 2y - 7 = 0 \]

5. **Rearrange to Standard Circle Form**:
Rearranging the equation:
\[ x^2 + 2x + y^2 - 2y = 7 \]
Now, complete the square for both \(x\) and \(y\):
- For \(x^2 + 2x\), we add and subtract \(1\):
\[ (x + 1)^2 - 1 \]
- For \(y^2 - 2y\), we add and subtract \(1\):
\[ (y - 1)^2 - 1 \]
Thus, the equation becomes:
\[ (x + 1)^2 - 1 + (y - 1)^2 - 1 = 7 \]
Simplifying gives:
\[ (x + 1)^2 + (y - 1)^2 = 9 \]

6. **Identify the Radius**:
The standard form of a circle is \((x - h)^2 + (y - k)^2 = r^2\), where \(r\) is the radius. Here, we have:
\[ r^2 = 9 \quad \Rightarrow \quad r = \sqrt{9} = 3 \]

### Final Answer:
The radius of the circle is \(3\).