The circles `x^(2)+y^(2)-4x-6y-12=0` and `x^(2)+y^(2)+4x+6y+4=0`

Options

A. touch externally
B. touch internally
C. intersect in two points
D. do not intersect

Answer

To solve the problem, we need to analyze the two given circles and determine their relationship (whether they touch externally, touch internally, intersect at two points, or do not intersect).

### Step 1: Write the equations of the circles in standard form

The first circle is given by the equation:
\[ x^2 + y^2 - 4x - 6y - 12 = 0 \]

We can rearrange this into standard form by completing the square.

1. Group the \(x\) and \(y\) terms:
\[ (x^2 - 4x) + (y^2 - 6y) = 12 \]

2. Complete the square for \(x\):
\[ x^2 - 4x = (x - 2)^2 - 4 \]

3. Complete the square for \(y\):
\[ y^2 - 6y = (y - 3)^2 - 9 \]

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

So, the first circle has center \(C_1(2, 3)\) and radius \(r_1 = 5\).

The second circle is given by the equation:
\[ x^2 + y^2 + 4x + 6y + 4 = 0 \]

Again, we rearrange this into standard form by completing the square.

1. Group the \(x\) and \(y\) terms:
\[ (x^2 + 4x) + (y^2 + 6y) = -4 \]

2. Complete the square for \(x\):
\[ x^2 + 4x = (x + 2)^2 - 4 \]

3. Complete the square for \(y\):
\[ y^2 + 6y = (y + 3)^2 - 9 \]

4. Substitute back into the equation:
\[ (x + 2)^2 - 4 + (y + 3)^2 - 9 = -4 \]
\[ (x + 2)^2 + (y + 3)^2 = 3 \]

So, the second circle has center \(C_2(-2, -3)\) and radius \(r_2 = \sqrt{3}\).

### Step 2: Calculate the distance between the centers of the circles

The distance \(d\) between the centers \(C_1(2, 3)\) and \(C_2(-2, -3)\) is given by the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{((-2) - 2)^2 + ((-3) - 3)^2} \]
\[ d = \sqrt{(-4)^2 + (-6)^2} \]
\[ d = \sqrt{16 + 36} \]
\[ d = \sqrt{52} = 2\sqrt{13} \]

### Step 3: Compare the distance with the sum and difference of the radii

Now we need to find \(r_1 + r_2\) and \(r_1 - r_2\):
- \(r_1 + r_2 = 5 + \sqrt{3}\)
- \(r_1 - r_2 = 5 - \sqrt{3}\)

### Step 4: Determine the relationship between the circles

1. If \(d > r_1 + r_2\): circles do not intersect.
2. If \(d = r_1 + r_2\): circles touch externally.
3. If \(d < r_1 + r_2\) and \(d > |r_1 - r_2|\): circles intersect at two points.
4. If \(d = |r_1 - r_2|\): circles touch internally.
5. If \(d < |r_1 - r_2|\): circles do not intersect.

Now, we need to evaluate \(d\):
- \(d = 2\sqrt{13} \approx 7.2\)
- \(r_1 + r_2 = 5 + \sqrt{3} \approx 5 + 1.73 = 6.73\)
- \(r_1 - r_2 = 5 - \sqrt{3} \approx 5 - 1.73 = 3.27\)

Since \(7.2 > 6.73\) and \(7.2 > 3.27\), we conclude that the circles intersect at two points.

### Final Answer:
The circles intersect at two points.