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

Options

A. cut orthogonally
B. touch each other internally
C. intersect in two points
D. touch each other externally

Answer

To solve the problem of determining the relationship between the two circles given by the equations \(x^2 + y^2 + 6x + 6y = 0\) and \(x^2 + y^2 - 12x - 12y = 0\), we will follow these steps:

### Step 1: Rewrite the equations in standard form
We start by rewriting each circle's equation in standard form \((x - h)^2 + (y - k)^2 = r^2\).

1. **First Circle:**
\[ x^2 + y^2 + 6x + 6y = 0 \]
Rearranging gives:
\[ x^2 + 6x + y^2 + 6y = 0 \]
Completing the square:
\[ (x^2 + 6x + 9) + (y^2 + 6y + 9) = 18 \]
This simplifies to:
\[ (x + 3)^2 + (y + 3)^2 = 18 \]
Thus, the center is \((-3, -3)\) and the radius \(r_1 = \sqrt{18} = 3\sqrt{2}\).

2. **Second Circle:**
\[ x^2 + y^2 - 12x - 12y = 0 \]
Rearranging gives:
\[ x^2 - 12x + y^2 - 12y = 0 \]
Completing the square:
\[ (x^2 - 12x + 36) + (y^2 - 12y + 36) = 72 \]
This simplifies to:
\[ (x - 6)^2 + (y - 6)^2 = 72 \]
Thus, the center is \((6, 6)\) and the radius \(r_2 = \sqrt{72} = 6\sqrt{2}\).

### Step 2: Calculate the distance between the centers
Next, we calculate the distance \(d\) between the centers of the two circles:
\[ d = \sqrt{(6 - (-3))^2 + (6 - (-3))^2} = \sqrt{(6 + 3)^2 + (6 + 3)^2} = \sqrt{9^2 + 9^2} = \sqrt{81 + 81} = \sqrt{162} = 9\sqrt{2} \]

### Step 3: Compare the distance with the sum of the radii
Now, we find the sum of the radii:
\[ r_1 + r_2 = 3\sqrt{2} + 6\sqrt{2} = 9\sqrt{2} \]

### Step 4: Determine the relationship between the circles
Since the distance between the centers \(d = 9\sqrt{2}\) is equal to the sum of the radii \(r_1 + r_2 = 9\sqrt{2}\), this indicates that the circles touch each other externally.

### Conclusion
The two circles touch each other externally.