If `A_(1), A_(2)`, be the areas of the curves
`C_(1)" "x^(2)+y^(2)+18x+24y=0`
`and c_(2)" "x^(2)/14+y^(2)/13=1` them

Options

A. `A_(1) gt A_(2)`
B. `A_(1) lt A_(2)`
C. `A_(1)=A_(2)`
D. none of these

Answer

To find the areas of the curves \( C_1 \) and \( C_2 \), we will analyze each curve step by step.

### Step 1: Analyze the first curve \( C_1 \)

The equation of the first curve is given by:
\[ x^2 + y^2 + 18x + 24y = 0 \]
This can be rewritten to identify the center and radius of the circle.

### Step 2: Complete the square for \( C_1 \)

We will complete the square for both \( x \) and \( y \).

1. For \( x \):
\[ x^2 + 18x = (x + 9)^2 - 81 \]
2. For \( y \):
\[ y^2 + 24y = (y + 12)^2 - 144 \]

Now substituting back into the equation:
\[ (x + 9)^2 - 81 + (y + 12)^2 - 144 = 0 \]
This simplifies to:
\[ (x + 9)^2 + (y + 12)^2 = 225 \]

### Step 3: Identify the center and radius of the circle

From the equation \( (x + 9)^2 + (y + 12)^2 = 15^2 \):
- The center of the circle \( C_1 \) is \( (-9, -12) \).
- The radius \( r_1 \) is \( 15 \).

### Step 4: Calculate the area of the circle \( C_1 \)

The area \( A_1 \) of a circle is given by the formula:
\[ A_1 = \pi r^2 \]
Substituting the radius:
\[ A_1 = \pi (15)^2 = 225\pi \]

### Step 5: Analyze the second curve \( C_2 \)

The equation of the second curve is given by:
\[ \frac{x^2}{14} + \frac{y^2}{13} = 1 \]
This represents an ellipse.

### Step 6: Identify the semi-major and semi-minor axes

Comparing with the standard form of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \):
- Here, \( a^2 = 14 \) and \( b^2 = 13 \).
- Thus, \( a = \sqrt{14} \) and \( b = \sqrt{13} \).

### Step 7: Calculate the area of the ellipse \( C_2 \)

The area \( A_2 \) of an ellipse is given by the formula:
\[ A_2 = \pi a b \]
Substituting the values of \( a \) and \( b \):
\[ A_2 = \pi \cdot \sqrt{14} \cdot \sqrt{13} = \pi \cdot \sqrt{182} \]

### Step 8: Compare the areas \( A_1 \) and \( A_2 \)

We have:
- \( A_1 = 225\pi \)
- \( A_2 = \pi \cdot \sqrt{182} \)

Since \( 225 > \sqrt{182} \), it follows that:
\[ A_1 > A_2 \]

### Conclusion

Thus, the area of the circle \( C_1 \) is greater than the area of the ellipse \( C_2 \).