Find the centre and radius of the circle `2x^(2)+2y^(2)=3x-5y+7`

Answer

To find the center and radius of the circle given by the equation \(2x^2 + 2y^2 = 3x - 5y + 7\), we will follow these steps:

### Step 1: Rearrange the equation
First, we need to rearrange the equation to bring all terms to one side:
\[ 2x^2 + 2y^2 - 3x + 5y - 7 = 0 \]

### Step 2: Divide by 2
Next, we divide the entire equation by 2 to simplify it:
\[ x^2 + y^2 - \frac{3}{2}x + \frac{5}{2}y - \frac{7}{2} = 0 \]

### Step 3: Group the terms
Now, we will group the \(x\) and \(y\) terms:
\[ x^2 - \frac{3}{2}x + y^2 + \frac{5}{2}y = \frac{7}{2} \]

### Step 4: Complete the square for \(x\)
To complete the square for the \(x\) terms:
1. Take the coefficient of \(x\) which is \(-\frac{3}{2}\), halve it to get \(-\frac{3}{4}\), and square it to get \(\frac{9}{16}\).
2. Add and subtract \(\frac{9}{16}\):
\[ \left(x - \frac{3}{4}\right)^2 - \frac{9}{16} \]

### Step 5: Complete the square for \(y\)
Now, for the \(y\) terms:
1. Take the coefficient of \(y\) which is \(\frac{5}{2}\), halve it to get \(\frac{5}{4}\), and square it to get \(\frac{25}{16}\).
2. Add and subtract \(\frac{25}{16}\):
\[ \left(y + \frac{5}{4}\right)^2 - \frac{25}{16} \]

### Step 6: Substitute back into the equation
Substituting back into the equation gives:
\[ \left(x - \frac{3}{4}\right)^2 - \frac{9}{16} + \left(y + \frac{5}{4}\right)^2 - \frac{25}{16} = \frac{7}{2} \]

### Step 7: Combine constants
Combine the constant terms:
\[ -\frac{9}{16} - \frac{25}{16} = -\frac{34}{16} = -\frac{17}{8} \]
So, we have:
\[ \left(x - \frac{3}{4}\right)^2 + \left(y + \frac{5}{4}\right)^2 = \frac{7}{2} + \frac{17}{8} \]
Finding a common denominator (which is 8):
\[ \frac{7}{2} = \frac{28}{8} \]
Thus:
\[ \frac{28}{8} + \frac{17}{8} = \frac{45}{8} \]

### Step 8: Final equation of the circle
Now, we can write the equation of the circle as:
\[ \left(x - \frac{3}{4}\right)^2 + \left(y + \frac{5}{4}\right)^2 = \frac{45}{8} \]

### Step 9: Identify the center and radius
From the standard form \((x - h)^2 + (y - k)^2 = r^2\):
- The center \((h, k)\) is \(\left(\frac{3}{4}, -\frac{5}{4}\right)\)
- The radius \(r\) is \(\sqrt{\frac{45}{8}} = \frac{3\sqrt{5}}{4}\)

### Final Answer
- **Center**: \(\left(\frac{3}{4}, -\frac{5}{4}\right)\)
- **Radius**: \(\frac{3\sqrt{5}}{4}\)