The locus of the centre of the circle which moves such that it touches the circle `(x + 1)^(2) + y^(2) = 1` externally and also the y-axis is

Options

A. `y^(2) = 4x, x ge 0`
B. `y^(2) = -4x, x le 0`
C. `y = 0, x gt 0`
D. `y = 0, AA x in R`

Answer

To find the locus of the center of a circle that touches the given circle \((x + 1)^2 + y^2 = 1\) externally and also touches the y-axis, we will follow these steps:

### Step 1: Identify the given circle's properties
The given equation \((x + 1)^2 + y^2 = 1\) represents a circle with:
- Center: \((-1, 0)\)
- Radius: \(1\)

### Step 2: Define the center of the moving circle
Let the center of the moving circle be denoted as \(C(h, k)\).

### Step 3: Condition for external tangency
For the moving circle to touch the given circle externally, the distance between the centers of the two circles must equal the sum of their radii. If we denote the radius of the moving circle as \(r\), then the condition is:
\[ \text{Distance}(C, (-1, 0)) = r + 1 \]
This can be expressed using the distance formula:
\[ \sqrt{(h + 1)^2 + k^2} = r + 1 \]

### Step 4: Condition for touching the y-axis
For the circle to touch the y-axis, the distance from the center \(C(h, k)\) to the y-axis must equal the radius \(r\):
\[ |h| = r \]

### Step 5: Substitute \(r\) from the y-axis condition
From the condition \(|h| = r\), we can express \(r\) as:
\[ r = |h| \]
Substituting this into the external tangency condition gives:
\[ \sqrt{(h + 1)^2 + k^2} = |h| + 1 \]

### Step 6: Square both sides to eliminate the square root
Squaring both sides leads to:
\[ (h + 1)^2 + k^2 = (|h| + 1)^2 \]

### Step 7: Expand both sides
Expanding both sides:
\[ (h^2 + 2h + 1 + k^2) = (h^2 + 2|h| + 1) \]

### Step 8: Simplify the equation
Canceling \(h^2 + 1\) from both sides:
\[ 2h + k^2 = 2|h| \]

### Step 9: Analyze the cases for \(h\)
We need to consider two cases for \(|h|\):
1. **Case 1:** \(h \geq 0\) (then \(|h| = h\)):
\[ 2h + k^2 = 2h \implies k^2 = 0 \implies k = 0 \]
So, for \(h \geq 0\), the locus is \(y = 0\) for \(x \geq 0\).

2. **Case 2:** \(h < 0\) (then \(|h| = -h\)):
\[ 2h + k^2 = -2h \implies 4h + k^2 = 0 \implies k^2 = -4h \]
Since \(h < 0\), \(k^2\) must be non-negative, which means \(h\) must be less than or equal to \(0\).

### Step 10: Conclusion
The locus of the center of the circle is:
- For \(h \geq 0\): \(y = 0\) (the x-axis) for \(x \geq 0\)
- For \(h < 0\): The equation \(k^2 = -4h\) describes a parabola that opens to the left.

Thus, the final answer is:
The locus of the center of the circle is the right half of the x-axis and the left parabola defined by \(k^2 = -4h\).