The locus of the centre of a circle of radius 2 which rolls on the outside of the circle `x^(2)+y^(2)+3x-6y-9=0` is

Options

A. `x^(2)+y^(2)+3x-6y+5=0`
B. `x^(2)+y^(2)+3x-6y-31=0 `
C. `x^(2)+y^(2)+3x-6y+29//4=0`
D. none

Answer

To find the locus of the center of a circle of radius 2 that rolls on the outside of the circle given by the equation \(x^2 + y^2 + 3x - 6y - 9 = 0\), we can follow these steps:

### Step 1: Identify the center and radius of the given circle
The equation of the circle can be rewritten in standard form. We start by completing the square for both \(x\) and \(y\).

1. Rearranging the equation:
\[ x^2 + 3x + y^2 - 6y = 9 \]

2. Completing the square for \(x\):
\[ x^2 + 3x = (x + \frac{3}{2})^2 - \frac{9}{4} \]

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

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

5. Simplifying gives:
\[ (x + \frac{3}{2})^2 + (y - 3)^2 = \frac{81}{4} \]

From this, we can identify:
- The center \(C(-\frac{3}{2}, 3)\)
- The radius \(r = \frac{9}{2}\)

### Step 2: Set up the relationship for the locus of the center of the rolling circle
Let the center of the rolling circle be \(C(h, k)\). The distance between the centers of the two circles must equal the sum of their radii since the rolling circle is outside the given circle.

The distance between the centers is given by:
\[ \sqrt{(h + \frac{3}{2})^2 + (k - 3)^2} \]
This distance must equal the sum of the radii:
\[ \frac{9}{2} + 2 = \frac{13}{2} \]

### Step 3: Set up the equation
Equating the distance to the sum of the radii:
\[ \sqrt{(h + \frac{3}{2})^2 + (k - 3)^2} = \frac{13}{2} \]

### Step 4: Square both sides
Squaring both sides gives:
\[ (h + \frac{3}{2})^2 + (k - 3)^2 = \left(\frac{13}{2}\right)^2 \]
\[ (h + \frac{3}{2})^2 + (k - 3)^2 = \frac{169}{4} \]

### Step 5: Rearranging to standard form
This can be rewritten as:
\[ (h + \frac{3}{2})^2 + (k - 3)^2 - \frac{169}{4} = 0 \]

### Step 6: Replace \(h\) and \(k\) with \(x\) and \(y\)
Let \(h = x\) and \(k = y\):
\[ (x + \frac{3}{2})^2 + (y - 3)^2 = \frac{169}{4} \]

### Step 7: Final equation
This represents a circle with center \((- \frac{3}{2}, 3)\) and radius \(\frac{13}{2}\).

### Conclusion
The locus of the center of the circle of radius 2 that rolls on the outside of the given circle is:
\[ (x + \frac{3}{2})^2 + (y - 3)^2 = \frac{169}{4} \]