Find the diameter of the circle `2x^(2)+2y^(2)-6x-9=0`.

Answer

To find the diameter of the circle given by the equation \(2x^2 + 2y^2 - 6x - 9 = 0\), we can follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the equation:
\[ 2x^2 + 2y^2 - 6x - 9 = 0 \]
To simplify, we can divide the entire equation by 2:
\[ x^2 + y^2 - 3x - \frac{9}{2} = 0 \]

### Step 2: Rearrange the equation
Next, we rearrange the equation to group the \(x\) and \(y\) terms:
\[ x^2 - 3x + y^2 = \frac{9}{2} \]

### Step 3: Complete the square for the \(x\) terms
To complete the square for the \(x\) terms, we take the coefficient of \(x\), which is \(-3\), halve it to get \(-\frac{3}{2}\), and square it:
\[ \left(-\frac{3}{2}\right)^2 = \frac{9}{4} \]
We add and subtract \(\frac{9}{4}\) inside the equation:
\[ \left(x^2 - 3x + \frac{9}{4}\right) + y^2 = \frac{9}{2} + \frac{9}{4} \]
This simplifies to:
\[ \left(x - \frac{3}{2}\right)^2 + y^2 = \frac{18}{4} + \frac{9}{4} = \frac{27}{4} \]

### Step 4: Identify the center and radius
Now we have the equation in standard form:
\[ \left(x - \frac{3}{2}\right)^2 + y^2 = \frac{27}{4} \]
From this, we can identify:
- The center of the circle is \(\left(\frac{3}{2}, 0\right)\)
- The radius \(r\) is given by:
\[ r = \sqrt{\frac{27}{4}} = \frac{3\sqrt{3}}{2} \]

### Step 5: Calculate the diameter
The diameter \(d\) of the circle is twice the radius:
\[ d = 2r = 2 \times \frac{3\sqrt{3}}{2} = 3\sqrt{3} \]

Thus, the diameter of the circle is:
\[ \boxed{3\sqrt{3}} \]

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