If the length of the daimeter of a circle is equal to the length of the major axis of the ellipse whose equation is `x^(2)+4y^(2)-4x+8y-28` , to the nearest whole number , what is the area of the cirle ?

Options

A. 28
B. 64
C. 113
D. 254

Explanation

To solve the problem step by step, we will first convert the given equation of the ellipse into its standard form, then determine the length of the major axis, and finally calculate the area of the circle.

### Step 1: Write the equation of the ellipse
The given equation of the ellipse is:
\[ x^2 + 4y^2 - 4x + 8y - 28 = 0 \]

### Step 2: Rearranging the equation
Rearranging the equation, we have:
\[ x^2 - 4x + 4y^2 + 8y = 28 \]

### Step 3: Completing the square for \(x\)
To complete the square for \(x\):
\[ x^2 - 4x = (x - 2)^2 - 4 \]

### Step 4: Completing the square for \(y\)
For the \(y\) terms, factor out 4:
\[ 4(y^2 + 2y) = 4((y + 1)^2 - 1) = 4(y + 1)^2 - 4 \]

### Step 5: Substitute back into the equation
Substituting back, we get:
\[ (x - 2)^2 - 4 + 4(y + 1)^2 - 4 = 28 \]
\[ (x - 2)^2 + 4(y + 1)^2 - 8 = 28 \]
\[ (x - 2)^2 + 4(y + 1)^2 = 36 \]

### Step 6: Divide by 36 to get standard form
Dividing the whole equation by 36:
\[ \frac{(x - 2)^2}{36} + \frac{(y + 1)^2}{9} = 1 \]

### Step 7: Identify the semi-major and semi-minor axes
From the standard form of the ellipse \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\):
- Here, \(a^2 = 36\) (so \(a = 6\)) and \(b^2 = 9\) (so \(b = 3\)).
- The length of the major axis is \(2a = 2 \times 6 = 12\).

### Step 8: Diameter of the circle
The diameter of the circle is equal to the length of the major axis of the ellipse, which is 12.

### Step 9: Calculate the radius of the circle
The radius \(r\) of the circle is:
\[ r = \frac{diameter}{2} = \frac{12}{2} = 6 \]

### Step 10: Calculate the area of the circle
The area \(A\) of the circle is given by the formula:
\[ A = \pi r^2 = \pi (6^2) = 36\pi \]

### Step 11: Approximate the area
Using \(\pi \approx 3.14\):
\[ A \approx 36 \times 3.14 = 113.04 \]

### Final Answer
To the nearest whole number, the area of the circle is:
\[ \boxed{113} \]