(i) Centre of the circle `x^(2) + y^(2) - 8x + 10 y + 12 = 0 ` is
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 centers of the given circles, we will follow the steps of completing the square for each equation.

### Step 1: Find the center of the first circle

The equation of the first circle is:
\[ x^2 + y^2 - 8x + 10y + 12 = 0 \]

**Rearranging the equation:**
\[ x^2 - 8x + y^2 + 10y + 12 = 0 \]

**Completing the square for \(x\):**
1. Take the coefficient of \(x\) which is \(-8\), halve it to get \(-4\), and square it to get \(16\).
2. Add and subtract \(16\) in the equation.

**Completing the square for \(y\):**
1. Take the coefficient of \(y\) which is \(10\), halve it to get \(5\), and square it to get \(25\).
2. Add and subtract \(25\) in the equation.

**Now, rewrite the equation:**
\[ (x^2 - 8x + 16) + (y^2 + 10y + 25) + 12 - 16 - 25 = 0 \]
This simplifies to:
\[ (x - 4)^2 + (y + 5)^2 - 29 = 0 \]
So, we can rewrite it as:
\[ (x - 4)^2 + (y + 5)^2 = 29 \]

**Identifying the center:**
From the equation \((x - h)^2 + (y - k)^2 = r^2\), we can see that the center \((h, k)\) is:
\[ (4, -5) \]

### Step 2: Find the center of the second circle

The equation of the second circle is:
\[ 2x^2 + 2y^2 - x = 0 \]

**Rearranging the equation:**
First, divide the entire equation by \(2\):
\[ x^2 + y^2 - \frac{1}{2}x = 0 \]

**Completing the square for \(x\):**
1. Take the coefficient of \(x\) which is \(-\frac{1}{2}\), halve it to get \(-\frac{1}{4}\), and square it to get \(\frac{1}{16}\).
2. Add and subtract \(\frac{1}{16}\) in the equation.

**Now, rewrite the equation:**
\[ \left(x^2 - \frac{1}{2}x + \frac{1}{16}\right) + y^2 = \frac{1}{16} \]
This simplifies to:
\[ \left(x - \frac{1}{4}\right)^2 + y^2 = 0 \]

**Identifying the center:**
From the equation \((x - h)^2 + (y - k)^2 = r^2\), we can see that the center \((h, k)\) is:
\[ \left(\frac{1}{4}, 0\right) \]

### Final Centers:
1. The center of the first circle is \((4, -5)\).
2. The center of the second circle is \(\left(\frac{1}{4}, 0\right)\).

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