Locus of centre of a circle of radius `2,` which rolls on the outside of circle `x^2 +y^2 + 3x-6y-9=0` is

Options

A. a) `x^(2)+y^(2)+3x-6y-5=0`
B. b) `x^(2)+y^(2)+3x-6y-31=0`
C. c) `x^(2)+y^(2)+3x-6y-11=0`
D. d) `x^(2)+y^(2)+3x-6y-36=0`

Answer

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

### Step 1: Rewrite the given circle equation in standard form
The given equation is:
\[ x^2 + y^2 + 3x - 6y - 9 = 0 \]
We can rearrange this to find the center and radius.

### Step 2: Identify the center and radius of the given circle
To convert the equation into standard form, we complete the square for both \(x\) and \(y\).

1. For \(x\):
\[ x^2 + 3x \quad \text{(Complete the square)} \]
\[ = (x + \frac{3}{2})^2 - \frac{9}{4} \]

2. For \(y\):
\[ y^2 - 6y \quad \text{(Complete the square)} \]
\[ = (y - 3)^2 - 9 \]

Putting it all together:
\[ (x + \frac{3}{2})^2 - \frac{9}{4} + (y - 3)^2 - 9 - 9 = 0 \]
\[ (x + \frac{3}{2})^2 + (y - 3)^2 = \frac{9}{4} + 9 + 9 \]
\[ = \frac{9}{4} + \frac{36}{4} + \frac{36}{4} = \frac{81}{4} \]

Thus, the center of the circle is:
\[ \left(-\frac{3}{2}, 3\right) \]
And the radius \(R\) is:
\[ R = \sqrt{\frac{81}{4}} = \frac{9}{2} \]

### Step 3: Set up the locus of the center of the rolling circle
Let the center of the rolling circle be \((H, K)\). The distance between the center of the given circle and the center of the rolling circle is equal to the sum of their radii:
\[ \text{Distance} = R + r = \frac{9}{2} + 2 = \frac{9}{2} + \frac{4}{2} = \frac{13}{2} \]

The distance formula gives us:
\[ \sqrt{(H + \frac{3}{2})^2 + (K - 3)^2} = \frac{13}{2} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides:
\[ (H + \frac{3}{2})^2 + (K - 3)^2 = \left(\frac{13}{2}\right)^2 \]
\[ (H + \frac{3}{2})^2 + (K - 3)^2 = \frac{169}{4} \]

### Step 5: Expand and simplify
Expanding the left side:
\[ (H^2 + 3H + \frac{9}{4}) + (K^2 - 6K + 9) = \frac{169}{4} \]
Combining terms:
\[ H^2 + K^2 + 3H - 6K + \frac{9}{4} + 9 = \frac{169}{4} \]
Convert \(9\) to quarters:
\[ H^2 + K^2 + 3H - 6K + \frac{9}{4} + \frac{36}{4} = \frac{169}{4} \]
\[ H^2 + K^2 + 3H - 6K + \frac{45}{4} = \frac{169}{4} \]

### Step 6: Rearranging to form the locus equation
Rearranging gives:
\[ H^2 + K^2 + 3H - 6K - \frac{169 - 45}{4} = 0 \]
\[ H^2 + K^2 + 3H - 6K - \frac{124}{4} = 0 \]
\[ H^2 + K^2 + 3H - 6K - 31 = 0 \]

### Step 7: Replace \(H\) and \(K\) with \(x\) and \(y\)
Thus, the locus of the center of the rolling circle is:
\[ x^2 + y^2 + 3x - 6y - 31 = 0 \]

### Final Answer
The correct option is:
\[ \text{Option B: } x^2 + y^2 + 3x - 6y - 31 = 0 \]