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

Options

A. `{(x, y):x^(2)+(y-1)^(2)=4} cup {(x,y): y lt 0}`
B. `{(x,y):x^(2)=4y} cup {(0, y) : y lt 0}`
C. `{(x,y) :x^(2) =y} cup {(0, y) : y lt 0}`
D. `{(x,y):x^(2)=4y} cup {(x,y) : y gt 0}`

Answer

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

### Step 1: Identify the given circle's properties
The equation of the given circle is \( x^2 + (y - 1)^2 = 1 \).
- The center of this circle is \( (0, 1) \).
- The radius of this circle is \( 1 \).

### Step 2: Define the center of the required circle
Let the center of the required circle be \( (h, k) \) and its radius be \( r \).

### Step 3: Set up the conditions for tangency
1. The required circle touches the given circle externally.
2. The required circle touches the x-axis.

From the second condition, since the circle touches the x-axis, the distance from the center \( (h, k) \) to the x-axis must equal the radius \( r \). Therefore, we have:
\[ r = k \]

### Step 4: Set up the distance condition for external tangency
The distance between the centers of the two circles must equal the sum of their radii. The distance \( d \) between the centers \( (0, 1) \) and \( (h, k) \) is given by:
\[ d = \sqrt{(h - 0)^2 + (k - 1)^2} = \sqrt{h^2 + (k - 1)^2} \]
For external tangency, we have:
\[ d = r + 1 \]
Substituting \( r = k \):
\[ \sqrt{h^2 + (k - 1)^2} = k + 1 \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ h^2 + (k - 1)^2 = (k + 1)^2 \]
Expanding both sides:
\[ h^2 + (k^2 - 2k + 1) = (k^2 + 2k + 1) \]
This simplifies to:
\[ h^2 - 2k + 1 = 2k \]
Rearranging gives:
\[ h^2 = 4k - 1 \]

### Step 6: Express the equation in terms of \( k \)
This equation \( h^2 = 4k - 1 \) can be rearranged to express \( k \) in terms of \( h \):
\[ k = \frac{h^2 + 1}{4} \]

### Step 7: Identify the locus
The equation \( k = \frac{h^2 + 1}{4} \) represents a parabola that opens upwards.

### Conclusion
The locus of the center of the circle that touches the given circle externally and also touches the x-axis is given by the equation:
\[ y = \frac{x^2 + 1}{4} \]