Which of the following is/are true ? The circles `x^(2)+y^(2)-6x+6y+9=0` and `x^(2)+y^(2)+6x+6y+9=0` are such that :

Options

A. They do not intersect
B. They touch each other
C. Their exterior common tangents are parallel
D. Their interior common tangents are perpendicular

Answer

To solve the problem, we need to analyze the two circles given by their equations:

1. Circle 1: \( x^2 + y^2 - 6x - 6y + 9 = 0 \)
2. Circle 2: \( x^2 + y^2 + 6x + 6y + 9 = 0 \)

### Step 1: Rewrite the equations in standard form

We will complete the square for both circles.

**Circle 1:**
\[ x^2 - 6x + y^2 - 6y + 9 = 0 \]
Completing the square:
\[ (x^2 - 6x + 9) + (y^2 - 6y + 9) = 0 + 9 + 9 \]
\[ (x - 3)^2 + (y - 3)^2 = 9 \]
This gives us Circle 1 with center \( (3, 3) \) and radius \( r_1 = 3 \).

**Circle 2:**
\[ x^2 + 6x + y^2 + 6y + 9 = 0 \]
Completing the square:
\[ (x^2 + 6x + 9) + (y^2 + 6y + 9) = 0 - 9 - 9 \]
\[ (x + 3)^2 + (y + 3)^2 = -18 \]
This indicates that Circle 2 does not exist in the real plane since the radius squared is negative.

### Step 2: Analyze the centers and radii

- **Circle 1** has center \( C_1(3, 3) \) and radius \( r_1 = 3 \).
- **Circle 2** does not exist (as shown above).

### Step 3: Determine the relationship between the circles

Since Circle 2 does not exist, we can conclude that:
- The circles do not intersect.
- There are no common tangents since one of the circles is not real.

### Conclusion

The only true statement regarding the circles is that they do not intersect.