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 given circle \( x^2 + y^2 + 3x - 6y - 9 = 0 \), we will follow these steps:

### Step 1: Identify the center and radius of the given circle
The given equation of the circle can be rewritten in standard form. We start with:
\[ x^2 + y^2 + 3x - 6y - 9 = 0 \]
To convert this into standard form, we complete the square for \(x\) and \(y\).

For \(x\):
\[ x^2 + 3x \rightarrow (x + \frac{3}{2})^2 - \frac{9}{4} \]

For \(y\):
\[ y^2 - 6y \rightarrow (y - 3)^2 - 9 \]

Substituting these back into the equation gives:
\[ (x + \frac{3}{2})^2 - \frac{9}{4} + (y - 3)^2 - 9 - 9 = 0 \]
\[ (x + \frac{3}{2})^2 + (y - 3)^2 - \frac{9}{4} - 18 = 0 \]
\[ (x + \frac{3}{2})^2 + (y - 3)^2 = \frac{81}{4} \]

Thus, the center of the given circle is \(C(-\frac{3}{2}, 3)\) and the radius \(R = \frac{9}{2}\).

### Step 2: Determine the locus of the center of the rolling circle
The center of the circle that rolls on the outside will be at a distance equal to the radius of the given circle plus the radius of the rolling circle. Therefore, the distance from the center \(C\) to the center \(P\) of the rolling circle is:
\[ CP = R + r = \frac{9}{2} + 2 = \frac{9}{2} + \frac{4}{2} = \frac{13}{2} \]

### Step 3: Set up the distance equation
Let the coordinates of the center \(P\) of the rolling circle be \((h, k)\). The distance from \(C\) to \(P\) can be expressed using the distance formula:
\[ \sqrt{(h + \frac{3}{2})^2 + (k - 3)^2} = \frac{13}{2} \]

### Step 4: Square both sides to eliminate the square root
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: Expand and rearrange the equation
Expanding the left-hand side:
\[ (h^2 + 3h + \frac{9}{4}) + (k^2 - 6k + 9) = \frac{169}{4} \]
Combining terms gives:
\[ h^2 + k^2 + 3h - 6k + \frac{9}{4} + 9 = \frac{169}{4} \]
\[ h^2 + k^2 + 3h - 6k + \frac{45}{4} = \frac{169}{4} \]
Subtracting \(\frac{45}{4}\) from both sides:
\[ h^2 + k^2 + 3h - 6k = \frac{124}{4} = 31 \]

### Step 6: Final equation of the locus
Thus, the locus of the center of the circle that rolls on the outside of the given circle is:
\[ x^2 + y^2 + 3x - 6y - 31 = 0 \]