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 solve the problem, we need to find the equation of a circle that is concentric to the given circle and has double the area of that circle. Here’s a step-by-step solution:

### Step 1: Rewrite the given circle equation
The given equation of the circle is:
\[ 2x^2 + 2y^2 - 3x + 6y + 2 = 0 \]
We can simplify this 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 original circle
The general form of the equation of a circle is:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From our equation, we can identify:
- \( g = -\frac{3}{4} \)
- \( f = \frac{3}{2} \)
- \( c = 1 \)

The center of the circle is given by the coordinates \((-g, -f)\):
- Center: \(\left(\frac{3}{4}, -\frac{3}{2}\right)\)

### Step 3: Calculate the radius of the original circle
The radius \( R \) of the circle can be calculated using the formula:
\[ R = \sqrt{g^2 + f^2 - c} \]
Substituting the values:
\[ R = \sqrt{\left(-\frac{3}{4}\right)^2 + \left(\frac{3}{2}\right)^2 - 1} \]
Calculating each term:
- \( \left(-\frac{3}{4}\right)^2 = \frac{9}{16} \)
- \( \left(\frac{3}{2}\right)^2 = \frac{9}{4} = \frac{36}{16} \)
- Thus, \( R = \sqrt{\frac{9}{16} + \frac{36}{16} - 1} = \sqrt{\frac{45}{16} - \frac{16}{16}} = \sqrt{\frac{29}{16}} = \frac{\sqrt{29}}{4} \)

### Step 4: Calculate the area of the original circle
The area \( A \) of the circle is given by:
\[ A = \pi R^2 = \pi \left(\frac{\sqrt{29}}{4}\right)^2 = \pi \cdot \frac{29}{16} \]

### Step 5: Determine the area of the new circle
The new circle has double the area of the original circle:
\[ A_{new} = 2A = 2 \cdot \frac{29\pi}{16} = \frac{29\pi}{8} \]

### Step 6: Find the radius of the new circle
Let the radius of the new circle be \( R_{new} \). The area of the new circle is:
\[ \pi R_{new}^2 = \frac{29\pi}{8} \]
Dividing both sides by \( \pi \):
\[ R_{new}^2 = \frac{29}{8} \]
Taking the square root:
\[ R_{new} = \sqrt{\frac{29}{8}} = \frac{\sqrt{29}}{2\sqrt{2}} \]

### Step 7: Write the equation of the new circle
The new circle is concentric with the same center \(\left(\frac{3}{4}, -\frac{3}{2}\right)\) and has radius \( R_{new} = \frac{\sqrt{29}}{2\sqrt{2}} \).

Using the standard form of the circle equation:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Where \( h = \frac{3}{4} \), \( k = -\frac{3}{2} \), and \( r^2 = \frac{29}{8} \):
\[ \left(x - \frac{3}{4}\right)^2 + \left(y + \frac{3}{2}\right)^2 = \frac{29}{8} \]

### Step 8: Expand the equation
Expanding the left side:
\[ \left(x^2 - \frac{3}{2}x + \frac{9}{16}\right) + \left(y^2 + 3y + \frac{9}{4}\right) = \frac{29}{8} \]

Combining terms:
\[ x^2 + y^2 - \frac{3}{2}x + 3y + \frac{9}{16} + \frac{9}{4} = \frac{29}{8} \]
Convert \(\frac{9}{4}\) to sixteenths:
\[ \frac{9}{4} = \frac{36}{16} \]
So:
\[ x^2 + y^2 - \frac{3}{2}x + 3y + \frac{45}{16} = \frac{29}{8} \]

### Step 9: Move all terms to one side
To get the final equation:
\[ 16x^2 + 16y^2 - 24x + 48y - 58 = 0 \]

### Final Answer
Thus, the equation of the new circle is:
\[ 16x^2 + 16y^2 - 24x + 48y - 58 = 0 \]