A circle `C_(1)` of radius 2 units rolls o the outerside of the circle `C_(2) : x^(2) + y^(2) + 4x = 0` touching it externally.
The locus of the centre of `C_(1)` is

Options

A. `x^(2) + y^(2) + 4y -12 = 0`
B. `x^(2) + y^(2) + 4x -12 = 0`
C. `x^(2) + y^(2) + 4x + 4y -4 = 0`
D. `x^(2) + y^(2) - 4x = 0`

Answer

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

### Step 1: Identify the equation of circle \( C_2 \)

The given equation of circle \( C_2 \) is:

\[ x^2 + y^2 + 4x = 0 \]

We can rewrite this in standard form by completing the square.

### Step 2: Complete the square

To complete the square for the \( x \) terms:

\[ x^2 + 4x = (x + 2)^2 - 4 \]

Thus, we can rewrite the equation as:

\[ (x + 2)^2 - 4 + y^2 = 0 \implies (x + 2)^2 + y^2 = 4 \]

This shows that circle \( C_2 \) has a center at \( (-2, 0) \) and a radius of \( 2 \) units.

### Step 3: Determine the radius of circle \( C_1 \)

Circle \( C_1 \) has a radius of \( 2 \) units. Since it rolls on the outside of circle \( C_2 \), the distance from the center of circle \( C_1 \) to the center of circle \( C_2 \) will be the sum of their radii.

### Step 4: Calculate the radius of the locus circle

The radius of the locus circle, which is the distance from the center of circle \( C_2 \) to the center of circle \( C_1 \), is:

\[ \text{Radius of } C_2 + \text{Radius of } C_1 = 2 + 2 = 4 \]

### Step 5: Write the equation of the locus circle

The center of the locus circle is the same as the center of circle \( C_2 \), which is \( (-2, 0) \). The equation of a circle in standard form is given by:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

where \( (h, k) \) is the center and \( r \) is the radius. Therefore, the equation of the locus circle is:

\[ (x + 2)^2 + y^2 = 4^2 \]

This simplifies to:

\[ (x + 2)^2 + y^2 = 16 \]

### Step 6: Rewrite in standard form

Expanding this gives:

\[ x^2 + 4x + 4 + y^2 = 16 \implies x^2 + y^2 + 4x - 12 = 0 \]

Thus, the final equation of the locus of the center of circle \( C_1 \) is:

\[ x^2 + y^2 + 4x - 12 = 0 \]