Centre of the circle `2x^(2) + 2y^(2) -x = 0 ` is :

Options

A. `((1)/(2) , 0)`
B. `(- (1)/(2) , 0)`
C. `((1)/(4) , 0)`
D. `(- (1)/(4) , 0)`

Answer

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

### Step 1: Simplify the Equation
Start with the given equation:
\[ 2x^2 + 2y^2 - x = 0 \]
Divide the entire equation by 2 to simplify:
\[ x^2 + y^2 - \frac{x}{2} = 0 \]

### Step 2: Rearrange the Equation
Rearranging gives:
\[ x^2 + y^2 = \frac{x}{2} \]

### Step 3: Complete the Square
To complete the square for the \(x\) terms, we rewrite the equation:
\[ x^2 - \frac{x}{2} + y^2 = 0 \]
Now, take the coefficient of \(x\) (which is \(-\frac{1}{2}\)), halve it to get \(-\frac{1}{4}\), and then square it to get \(\frac{1}{16}\). We can add and subtract this value:
\[ \left(x^2 - \frac{x}{2} + \frac{1}{16}\right) + y^2 = \frac{1}{16} \]

### Step 4: Rewrite the Equation
Now, we can rewrite the left side as a perfect square:
\[ \left(x - \frac{1}{4}\right)^2 + y^2 = \frac{1}{16} \]

### Step 5: Identify the Center and Radius
The equation is now in the standard form of a circle:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center and \(r\) is the radius. From our equation:
- \(h = \frac{1}{4}\)
- \(k = 0\)

Thus, the center of the circle is:
\[ \left(\frac{1}{4}, 0\right) \]

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

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