Find the centre and radius of the sphere `2x^2+2y^2+2z^2-2x-4y+2z+3=0`.

Answer

To find the center and radius of the sphere given by the equation \(2x^2 + 2y^2 + 2z^2 - 2x - 4y + 2z + 3 = 0\), we will follow these steps:

### Step 1: Simplify the Equation
First, we divide the entire equation by 2 to make the coefficients of \(x^2\), \(y^2\), and \(z^2\) equal to 1.

\[ x^2 + y^2 + z^2 - x - 2y + z + \frac{3}{2} = 0 \]

### Step 2: Rearrange the Equation
Next, we rearrange the equation to isolate the constant term on one side.

\[ x^2 + y^2 + z^2 - x - 2y + z = -\frac{3}{2} \]

### Step 3: Complete the Square
Now, we will complete the square for each variable \(x\), \(y\), and \(z\).

1. **For \(x\)**:
- The expression is \(x^2 - x\).
- Completing the square:
\[ x^2 - x = \left(x - \frac{1}{2}\right)^2 - \frac{1}{4} \]

2. **For \(y\)**:
- The expression is \(y^2 - 2y\).
- Completing the square:
\[ y^2 - 2y = (y - 1)^2 - 1 \]

3. **For \(z\)**:
- The expression is \(z^2 + z\).
- Completing the square:
\[ z^2 + z = \left(z + \frac{1}{2}\right)^2 - \frac{1}{4} \]

### Step 4: Substitute Back
Substituting these completed squares back into the equation gives:

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

### Step 5: Simplify the Equation
Now, we simplify the equation:

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

This simplifies to:

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

Adding \(\frac{3}{2}\) to both sides results in:

\[ \left(x - \frac{1}{2}\right)^2 + (y - 1)^2 + \left(z + \frac{1}{2}\right)^2 = 0 \]

### Step 6: Identify the Center and Radius
From the equation \(\left(x - \frac{1}{2}\right)^2 + (y - 1)^2 + \left(z + \frac{1}{2}\right)^2 = 0\), we can see that:

- The center of the sphere \((x_1, y_1, z_1)\) is \(\left(\frac{1}{2}, 1, -\frac{1}{2}\right)\).
- The radius \(r\) is \(0\) (since the equation equals zero).

### Final Answer
- **Center**: \(\left(\frac{1}{2}, 1, -\frac{1}{2}\right)\)
- **Radius**: \(0\)