The locus of the centres of circles which touches `(y-1)^(2)+x^(2)=1` externally and also touches X-axis is

Options

A. `{x^(2)=4y,yge0}uu{0,y),ylt}`
B. `x^(2)=y`
C. `y=4x^(2)`
D. `y^(2)=4x uu(0,y),yinR`

Explanation

To solve the problem of finding the locus of the centers of circles that touch the circle \((y-1)^2 + x^2 = 1\) externally and also touch the X-axis, we can follow these steps:

### Step 1: Understand the given circle
The equation \((y-1)^2 + x^2 = 1\) represents a circle centered at \((0, 1)\) with a radius of \(1\).

### Step 2: Set up the center of the new circle
Let the center of the new circle be \((H, K)\) and let its radius be \(R\). Since the new circle touches the X-axis, the radius \(R\) will be equal to the perpendicular distance from the center to the X-axis, which is \(|K|\). Therefore, we have:
\[ R = |K| \]

### Step 3: Use the condition for external tangency
The distance between the centers of the two circles must equal the sum of their radii. The distance between the centers \((0, 1)\) and \((H, K)\) can be calculated using the distance formula:
\[ \sqrt{(H - 0)^2 + (K - 1)^2} = \sqrt{H^2 + (K - 1)^2} \]
Since the new circle touches the given circle externally, we have:
\[ \sqrt{H^2 + (K - 1)^2} = R + 1 \]
Substituting \(R = |K|\) into the equation gives:
\[ \sqrt{H^2 + (K - 1)^2} = |K| + 1 \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides results in:
\[ H^2 + (K - 1)^2 = (|K| + 1)^2 \]
Expanding both sides:
\[ H^2 + (K^2 - 2K + 1) = (K^2 + 2|K| + 1) \]
This simplifies to:
\[ H^2 - 2K = 2|K| \]

### Step 5: Consider two cases for \(K\)
**Case 1:** \(K \geq 0\) (where \(|K| = K\)):
\[ H^2 - 2K = 2K \implies H^2 = 4K \implies K = \frac{H^2}{4} \]

**Case 2:** \(K < 0\) (where \(|K| = -K\)):
\[ H^2 - 2K = -2K \implies H^2 = 0 \implies H = 0 \]
In this case, \(K\) can be any negative value.

### Step 6: Combine the results
From Case 1, we have the equation of the parabola:
\[ H^2 = 4K \quad \text{for } K \geq 0 \]
From Case 2, we have the line:
\[ H = 0 \quad \text{for } K < 0 \]

### Final Result
The locus of the centers of the circles is given by:
1. The parabola \(x^2 = 4y\) for \(y \geq 0\).
2. The vertical line \(x = 0\) for \(y < 0\).