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 can simplify the equation by dividing all terms by 2:

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

### Step 2: Rearrange the Equation
Rearranging gives us:

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

### Step 3: Complete the Square for Each Variable
Next, we will complete the square for the \(x\), \(y\), and \(z\) terms.

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

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

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

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

\[ \left( x - \frac{1}{2} \right)^2 - \frac{1}{4} + \left( y - 1 \right)^2 - 1 + \left( z + \frac{1}{2} \right)^2 - \frac{1}{4} + \frac{3}{2} = 0. \]

### Step 5: Combine and Simplify
Now, combine the constants:

\[ \left( x - \frac{1}{2} \right)^2 + \left( y - 1 \right)^2 + \left( z + \frac{1}{2} \right)^2 - \frac{1}{4} - 1 - \frac{1}{4} + \frac{3}{2} = 0. \]

Calculating the constants:

\[ -\frac{1}{4} - 1 - \frac{1}{4} + \frac{3}{2} = -\frac{1}{4} - \frac{4}{4} - \frac{1}{4} + \frac{6}{4} = 0. \]

Thus, we have:

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

### Step 6: Identify the Center and Radius
From the equation of the sphere:

\[ (x - a)^2 + (y - b)^2 + (z - c)^2 = r^2, \]

we can see that:

- The center \((a, b, c)\) is \(\left(\frac{1}{2}, 1, -\frac{1}{2}\right)\).
- The radius \(r\) is \(0\) (since the right side equals \(0\)).

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

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