A circle C touches the x-axis and the circle `x ^(2) + (y-1) ^(2) =1` externally, then locus of the centre of the circle C is given by

Options

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

Answer

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

### Step 1: Understand the given circles
The circle defined by \( x^2 + (y-1)^2 = 1 \) has its center at \( (0, 1) \) and a radius of \( 1 \). The circle \( C \) touches the x-axis, meaning its center must be at a point \( (x, y) \) where \( y \) is equal to its radius \( r \).

### Step 2: Set up the relationship between the circles
Let the center of circle \( C \) be \( (x, y) \) and its radius be \( r \). Since circle \( C \) touches the x-axis, we have:
\[ y = r \]

### Step 3: Use the external tangency condition
For two circles to touch externally, the distance between their centers must equal the sum of their radii. The distance between the center of circle \( C \) at \( (x, y) \) and the center of the given circle \( (0, 1) \) is given by:
\[ \sqrt{(x - 0)^2 + (y - 1)^2} = \sqrt{x^2 + (y - 1)^2} \]
The sum of the radii is:
\[ r + 1 \]
Substituting \( r = y \) (from Step 1), we have:
\[ \sqrt{x^2 + (y - 1)^2} = y + 1 \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:
\[ x^2 + (y - 1)^2 = (y + 1)^2 \]

### Step 5: Expand both sides
Expanding both sides results in:
\[ x^2 + (y^2 - 2y + 1) = (y^2 + 2y + 1) \]
This simplifies to:
\[ x^2 - 2y + 1 = 2y \]

### Step 6: Rearranging the equation
Rearranging the equation gives:
\[ x^2 = 4y - 1 \]

### Step 7: Express the locus
This equation \( x^2 = 4y - 1 \) can be rewritten as:
\[ x^2 = 4(y - \frac{1}{4}) \]
This represents a parabola that opens upwards with its vertex at \( (0, \frac{1}{4}) \).

### Step 8: Consider the condition for \( y \)
Since the circle \( C \) touches the x-axis, \( y \) must be non-negative. Therefore, the locus of the center of circle \( C \) is given by:
\[ y \geq 0 \]

### Final Answer
The locus of the center of circle \( C \) is given by the parabola:
\[ x^2 = 4y - 1 \quad \text{for } y \geq 0 \]