Locus of the centre of circle of radius 2 which rolls on out side the rim of the circle `x^(2)+y^(2)-4x-6y-12=0` is

Options

A. a) `x^(2)+y^(2)-4x-6y=0`
B. b) `x^(2)+y^(2)-4x-6y-36=0`
C. c) `x^(2)+y^(2)-4x-6y+3=0`
D. d) `x^(2)+y^(2)-4x-6y-25=0`

Answer

To find the locus of the center of a circle of radius 2 that rolls on the outside of the given circle \(x^2 + y^2 - 4x - 6y - 12 = 0\), we will follow these steps:

### Step 1: Identify the center and radius of the given circle
The equation of the circle is given in the standard form:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
Here, we can identify:
- \(g = -2\)
- \(f = -3\)
- \(c = -12\)

The center \((h, k)\) of the circle can be found using the formulas:
\[ h = -g = 2, \quad k = -f = 3 \]
Thus, the center of the given circle is \((2, 3)\).

### Step 2: Calculate the radius of the given circle
The radius \(r\) of the circle can be calculated using the formula:
\[ r = \sqrt{g^2 + f^2 - c} \]
Substituting the values:
\[ r = \sqrt{(-2)^2 + (-3)^2 - (-12)} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5 \]
So, the radius of the given circle is 5.

### Step 3: Determine the distance from the center of the rolling circle to the center of the given circle
The radius of the rolling circle is 2. Therefore, the distance from the center of the rolling circle to the center of the given circle is:
\[ \text{Distance} = \text{Radius of given circle} + \text{Radius of rolling circle} = 5 + 2 = 7 \]

### Step 4: Set up the distance formula
Let \((h, k)\) be the center of the rolling circle. The distance between the point \((h, k)\) and the center of the given circle \((2, 3)\) can be expressed using the distance formula:
\[ \sqrt{(h - 2)^2 + (k - 3)^2} = 7 \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (h - 2)^2 + (k - 3)^2 = 49 \]

### Step 6: Expand the equation
Expanding the left-hand side:
\[ (h^2 - 4h + 4) + (k^2 - 6k + 9) = 49 \]
Combining like terms:
\[ h^2 + k^2 - 4h - 6k + 13 = 49 \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ h^2 + k^2 - 4h - 6k - 36 = 0 \]

### Step 8: Replace \(h\) and \(k\) with \(x\) and \(y\)
To find the locus, we replace \(h\) with \(x\) and \(k\) with \(y\):
\[ x^2 + y^2 - 4x - 6y - 36 = 0 \]

### Final Answer
Thus, the locus of the center of the circle of radius 2 that rolls on the outside of the given circle is:
\[ x^2 + y^2 - 4x - 6y - 36 = 0 \]