Centre and radius of circle ` 2x^(2) + 2y^(2) - 6x + 4y - 3 = 0 ` are

Options

A. `(-(2)/(3),1), (sqrt(19))/(2)`
B. `((2)/(3),-1), (sqrt(19))/(2)`
C. `((-3)/(2),-1), (sqrt(19))/(2)`
D. `((2)/(3),1), (sqrt(11))/(2)`

Answer

To find the center and radius of the circle given by the equation \( 2x^{2} + 2y^{2} - 6x + 4y - 3 = 0 \), we will follow these steps:

### Step 1: Rewrite the equation in standard form
The standard form of a circle's equation is:
\[ x^{2} + y^{2} + 2gx + 2fy + c = 0 \]
To convert the given equation into this form, we first divide the entire equation by 2:
\[ x^{2} + y^{2} - 3x + 2y - \frac{3}{2} = 0 \]

### Step 2: Identify coefficients
From the equation \( x^{2} + y^{2} - 3x + 2y - \frac{3}{2} = 0 \), we can identify:
- \( g = -\frac{3}{2} \)
- \( f = 1 \)
- \( c = -\frac{3}{2} \)

### Step 3: Find the center
The center of the circle can be found using the formula:
\[ \text{Center} = (-g, -f) \]
Substituting the values of \( g \) and \( f \):
\[ \text{Center} = \left(-\left(-\frac{3}{2}\right), -1\right) = \left(\frac{3}{2}, -1\right) \]

### Step 4: Find the radius
The radius can be calculated using the formula:
\[ \text{Radius} = \sqrt{g^{2} + f^{2} - c} \]
Substituting the values of \( g \), \( f \), and \( c \):
\[ \text{Radius} = \sqrt{\left(-\frac{3}{2}\right)^{2} + (1)^{2} - \left(-\frac{3}{2}\right)} \]
Calculating \( g^{2} \):
\[ \left(-\frac{3}{2}\right)^{2} = \frac{9}{4} \]
Calculating \( f^{2} \):
\[ (1)^{2} = 1 \]
Now substituting these values:
\[ \text{Radius} = \sqrt{\frac{9}{4} + 1 + \frac{3}{2}} \]
Converting \( 1 \) and \( \frac{3}{2} \) to have a common denominator of 4:
\[ 1 = \frac{4}{4}, \quad \frac{3}{2} = \frac{6}{4} \]
Now substituting:
\[ \text{Radius} = \sqrt{\frac{9}{4} + \frac{4}{4} + \frac{6}{4}} = \sqrt{\frac{19}{4}} = \frac{\sqrt{19}}{2} \]

### Final Answer
The center of the circle is \( \left(\frac{3}{2}, -1\right) \) and the radius is \( \frac{\sqrt{19}}{2} \).

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