If `A_(1),A_(2),A_(3)` be the areas of circles `x^(2)+y^(2)+4x+6y-19=0, x^(2)+y^(2)=9, x^(2)+y^(2)-4x-6y-12=0` respectively then

Options

A. `A_(1)gtA_(2)gtA_(3)`
B. `A_(2)gtA_(3)gtA_(1)`
C. `A_(1)gtA_(3)gtA_(2)`
D. `A_(2)gtA_(1)gtA_(3)`

Answer

To find the areas of the circles defined by the equations given in the question, we will first convert each equation into the standard form of a circle and then calculate the radius for each circle. The area of a circle is given by the formula \( A = \pi r^2 \), so we can compare the areas by comparing the radii.

### Step 1: Convert the first equation into standard form
The first circle is given by the equation:
\[ x^2 + y^2 + 4x + 6y - 19 = 0 \]

To convert this into standard form, we complete the square for the \( x \) and \( y \) terms.

1. Rearranging the equation:
\[ x^2 + 4x + y^2 + 6y = 19 \]

2. Completing the square:
- For \( x^2 + 4x \):
\[ x^2 + 4x = (x + 2)^2 - 4 \]
- For \( y^2 + 6y \):
\[ y^2 + 6y = (y + 3)^2 - 9 \]

3. Substituting back:
\[ (x + 2)^2 - 4 + (y + 3)^2 - 9 = 19 \]
\[ (x + 2)^2 + (y + 3)^2 = 32 \]

So, the center is \( (-2, -3) \) and the radius \( r_1 = \sqrt{32} = 4\sqrt{2} \).

### Step 2: Calculate the area of the first circle
Using the radius calculated:
\[ A_1 = \pi r_1^2 = \pi (4\sqrt{2})^2 = \pi \cdot 32 = 32\pi \]

### Step 3: Convert the second equation into standard form
The second circle is given by:
\[ x^2 + y^2 = 9 \]

This is already in standard form, where the center is \( (0, 0) \) and the radius \( r_2 = \sqrt{9} = 3 \).

### Step 4: Calculate the area of the second circle
Using the radius:
\[ A_2 = \pi r_2^2 = \pi (3)^2 = 9\pi \]

### Step 5: Convert the third equation into standard form
The third circle is given by:
\[ x^2 + y^2 - 4x - 6y - 12 = 0 \]

Completing the square:
1. Rearranging:
\[ x^2 - 4x + y^2 - 6y = 12 \]

2. Completing the square:
- For \( x^2 - 4x \):
\[ x^2 - 4x = (x - 2)^2 - 4 \]
- For \( y^2 - 6y \):
\[ y^2 - 6y = (y - 3)^2 - 9 \]

3. Substituting back:
\[ (x - 2)^2 - 4 + (y - 3)^2 - 9 = 12 \]
\[ (x - 2)^2 + (y - 3)^2 = 25 \]

So, the center is \( (2, 3) \) and the radius \( r_3 = \sqrt{25} = 5 \).

### Step 6: Calculate the area of the third circle
Using the radius:
\[ A_3 = \pi r_3^2 = \pi (5)^2 = 25\pi \]

### Step 7: Compare the areas
Now we have:
- \( A_1 = 32\pi \)
- \( A_2 = 9\pi \)
- \( A_3 = 25\pi \)

Comparing these areas:
- \( A_1 > A_3 > A_2 \)

### Final Result
Thus, the order of the areas is:
1. \( A_1 \) (largest)
2. \( A_3 \)
3. \( A_2 \)