The radius of the circle
`3x^(2)+3y^(2) +lambda xy +9x +(lambda-6)y+3=0`

Options

A. `3//2`
B. `2//3`
C. `(1)/(2)sqrt(17)`
D. none

Answer

To find the radius of the circle given by the equation:

\[ 3x^2 + 3y^2 + \lambda xy + 9x + (\lambda - 6)y + 3 = 0 \]

we will follow these steps:

### Step 1: Rewrite the equation in standard form
First, we divide the entire equation by 3 to simplify it:

\[ x^2 + y^2 + \frac{\lambda}{3}xy + 3x + \left(\frac{\lambda - 6}{3}\right)y + 1 = 0 \]

### Step 2: Identify the coefficients
In the general form of a circle \( x^2 + y^2 + Gx + Fy + C = 0 \), we identify:

- \( G = 3 \)
- \( F = \frac{\lambda - 6}{3} \)
- \( C = 1 \)

### Step 3: Apply the condition for a circle
For the equation to represent a circle, the following conditions must be satisfied:

1. The coefficients of \( x^2 \) and \( y^2 \) must be equal.
2. The coefficient of \( xy \) must be zero.

From the equation, we see that the coefficient of \( xy \) is \( \frac{\lambda}{3} \). Therefore, for the equation to represent a circle:

\[ \frac{\lambda}{3} = 0 \implies \lambda = 0 \]

### Step 4: Substitute \( \lambda \) back into the equation
Substituting \( \lambda = 0 \) into the equation gives:

\[ 3x^2 + 3y^2 + 9x - 6y + 3 = 0 \]

Dividing by 3:

\[ x^2 + y^2 + 3x - 2y + 1 = 0 \]

### Step 5: Rearranging to standard form
Rearranging the equation:

\[ x^2 + y^2 + 3x - 2y = -1 \]

### Step 6: Completing the square
Now, we complete the square for \( x \) and \( y \):

1. For \( x^2 + 3x \):
\[ x^2 + 3x = (x + \frac{3}{2})^2 - \frac{9}{4} \]

2. For \( y^2 - 2y \):
\[ y^2 - 2y = (y - 1)^2 - 1 \]

Substituting back:

\[ \left(x + \frac{3}{2}\right)^2 - \frac{9}{4} + (y - 1)^2 - 1 = -1 \]

This simplifies to:

\[ \left(x + \frac{3}{2}\right)^2 + (y - 1)^2 = \frac{9}{4} \]

### Step 7: Finding the radius
The equation is now in the standard form of a circle:

\[ \left(x - h\right)^2 + \left(y - k\right)^2 = r^2 \]

where \( r^2 = \frac{9}{4} \). Thus, the radius \( r \) is:

\[ r = \sqrt{\frac{9}{4}} = \frac{3}{2} \]

### Final Answer
The radius of the circle is:

\[ \boxed{\frac{3}{2}} \]