The equation of a circle `C_1` is `x^2+y^2-4x-2y-11=0` A circle`C_2` of radius `1` rolls on the outside of the circle `C_1` The locus of the centre `C_2` has the equation

Options

A. `x^(2)+y^(2)-4x-2y-20=0`
B. `x^(2)+y^(2)+4x+2y-20=0`
C. `x^(2)+y^(2)-3x-y-11=0`
D. none of these

Explanation

To find the locus of the center of circle \( C_2 \) that rolls on the outside of circle \( C_1 \), we will follow these steps:

### Step 1: Write the equation of circle \( C_1 \)
The equation of circle \( C_1 \) is given as:
\[ x^2 + y^2 - 4x - 2y - 11 = 0 \]

### Step 2: Convert the equation to standard form
To convert this to standard form, we complete the square for both \( x \) and \( y \).

1. Rearranging the equation:
\[ x^2 - 4x + y^2 - 2y = 11 \]

2. Completing the square for \( x \):
\[ x^2 - 4x = (x - 2)^2 - 4 \]

3. Completing the square for \( y \):
\[ y^2 - 2y = (y - 1)^2 - 1 \]

4. Substitute back into the equation:
\[ (x - 2)^2 - 4 + (y - 1)^2 - 1 = 11 \]
\[ (x - 2)^2 + (y - 1)^2 = 16 \]

This shows that circle \( C_1 \) has a center at \( (2, 1) \) and a radius \( r_1 = 4 \).

### Step 3: Determine the radius of circle \( C_2 \)
Circle \( C_2 \) has a radius \( r_2 = 1 \).

### Step 4: Set up the distance relation
The distance between the center of circle \( C_1 \) (let's call it \( C_1(2, 1) \)) and the center of circle \( C_2 \) (let's denote it as \( C_2(h, k) \)) must equal the sum of the radii of the two circles:
\[ \text{Distance}(C_1, C_2) = r_1 + r_2 = 4 + 1 = 5 \]

Using the distance formula:
\[ \sqrt{(h - 2)^2 + (k - 1)^2} = 5 \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (h - 2)^2 + (k - 1)^2 = 25 \]

### Step 6: Expand the equation
Expanding the left side:
\[ (h - 2)^2 = h^2 - 4h + 4 \]
\[ (k - 1)^2 = k^2 - 2k + 1 \]
So,
\[ h^2 - 4h + 4 + k^2 - 2k + 1 = 25 \]
Combining terms:
\[ h^2 + k^2 - 4h - 2k + 5 = 25 \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ h^2 + k^2 - 4h - 2k - 20 = 0 \]

### Step 8: Substitute \( h \) and \( k \) with \( x \) and \( y \)
Since \( h \) and \( k \) represent the coordinates of the center of circle \( C_2 \), we can substitute:
\[ x^2 + y^2 - 4x - 2y - 20 = 0 \]

### Final Answer
Thus, the locus of the center \( C_2 \) is given by:
\[ \boxed{x^2 + y^2 - 4x - 2y - 20 = 0} \]