The equation of the circle concentric to the circle `2x^(2)+2y^(2)-3x+6y+2=0` and having double the area of this circle, is

Options

A. `8x^(2)+8y^(2)-24x+48y-13=0`
B. `16x^(2)+16y^(2)+24x-48y-13=0`
C. `16x^(2)+16y^(2)-24x+48y-13=0`
D. `8x^(2)+8y^(2)+24x-48y-13=0`

Answer

To find the equation of the circle that is concentric to the given circle \(2x^2 + 2y^2 - 3x + 6y + 2 = 0\) and has double the area, we can follow these steps:

### Step 1: Simplify the given circle equation
First, we simplify the equation of the circle by dividing the entire equation by 2:

\[ x^2 + y^2 - \frac{3}{2}x + 3y + 1 = 0 \]

### Step 2: Identify the center and radius of the given circle
The standard form of a circle is given by:

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

From our simplified equation, we can identify:
- \(2g = -\frac{3}{2} \Rightarrow g = -\frac{3}{4}\)
- \(2f = 3 \Rightarrow f = \frac{3}{2}\)
- \(c = 1\)

The center of the circle \((h, k)\) is given by \((-g, -f)\):

\[ \text{Center} = \left(\frac{3}{4}, -\frac{3}{2}\right) \]

### Step 3: Calculate the radius of the given circle
The radius \(r\) can be calculated using the formula:

\[ r = \sqrt{g^2 + f^2 - c} \]

Substituting the values of \(g\), \(f\), and \(c\):

\[ r = \sqrt{\left(-\frac{3}{4}\right)^2 + \left(\frac{3}{2}\right)^2 - 1} \]

Calculating each term:

\[ = \sqrt{\frac{9}{16} + \frac{9}{4} - 1} \]

Converting \(\frac{9}{4}\) to have a common denominator of 16:

\[ = \sqrt{\frac{9}{16} + \frac{36}{16} - \frac{16}{16}} = \sqrt{\frac{29}{16}} = \frac{\sqrt{29}}{4} \]

### Step 4: Determine the radius of the new circle
The area of the original circle is given by:

\[ \text{Area} = \pi r^2 = \pi \left(\frac{\sqrt{29}}{4}\right)^2 = \frac{29\pi}{16} \]

Since we need a circle with double the area:

\[ \text{New Area} = 2 \times \frac{29\pi}{16} = \frac{29\pi}{8} \]

Let \(R\) be the radius of the new circle. Then:

\[ \pi R^2 = \frac{29\pi}{8} \]

Dividing both sides by \(\pi\):

\[ R^2 = \frac{29}{8} \]

### Step 5: Write the equation of the new circle
The equation of a circle with center \((h, k)\) and radius \(R\) is given by:

\[ (x - h)^2 + (y - k)^2 = R^2 \]

Substituting the center \(\left(\frac{3}{4}, -\frac{3}{2}\right)\) and \(R^2 = \frac{29}{8}\):

\[ \left(x - \frac{3}{4}\right)^2 + \left(y + \frac{3}{2}\right)^2 = \frac{29}{8} \]

### Step 6: Expand the equation
Expanding this equation:

\[ \left(x^2 - \frac{3}{2}x + \frac{9}{16}\right) + \left(y^2 + 3y + \frac{9}{4}\right) = \frac{29}{8} \]

Combining terms and moving everything to one side:

\[ x^2 + y^2 - \frac{3}{2}x + 3y + \left(\frac{9}{16} + \frac{9}{4} - \frac{29}{8}\right) = 0 \]

Finding a common denominator (16):

\[ \frac{9}{16} + \frac{36}{16} - \frac{58}{16} = \frac{45 - 58}{16} = -\frac{13}{16} \]

Thus, the final equation of the new circle is:

\[ x^2 + y^2 - \frac{3}{2}x + 3y - \frac{13}{16} = 0 \]

### Final Answer
The equation of the circle concentric to the given circle and having double the area is:

\[ 16x^2 + 16y^2 - 24x + 48y - 13 = 0 \]