Find the centre and radius of the circle `2x^(2) + 2y^(2) - x = 0`

Answer

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

### Step 1: Rewrite the equation
Start with the given equation:
\[ 2x^2 + 2y^2 - x = 0 \]
We can factor out a 2 from the terms involving \(x\) and \(y\):
\[ 2(x^2 + y^2) - x = 0 \]
Now, rearranging gives:
\[ 2(x^2 + y^2) = x \]
Dividing both sides by 2:
\[ x^2 + y^2 = \frac{x}{2} \]

### Step 2: Rearrange to standard form
Next, we want to rearrange this equation into the standard form of a circle. We can rewrite it as:
\[ x^2 - \frac{x}{2} + y^2 = 0 \]

### Step 3: Complete the square for \(x\)
To complete the square for the \(x\) terms, we take the coefficient of \(x\), which is \(-\frac{1}{2}\), halve it to get \(-\frac{1}{4}\), and square it to get \(\frac{1}{16}\). We add and subtract this value:
\[ x^2 - \frac{x}{2} + \frac{1}{16} + y^2 = \frac{1}{16} \]
This simplifies to:
\[ \left(x - \frac{1}{4}\right)^2 + y^2 = \frac{1}{16} \]

### Step 4: Identify the center and radius
Now, we can compare this equation with the standard form of a circle:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
From our equation, we see that:
- \(h = \frac{1}{4}\)
- \(k = 0\)
- \(r^2 = \frac{1}{16}\), thus \(r = \frac{1}{4}\)

### Conclusion
The center of the circle is \(\left(\frac{1}{4}, 0\right)\) and the radius is \(\frac{1}{4}\).