Find the equation of the circle which touches the x-axis and whose center is (1, 2).

Options

A. `x^(2)+y^(2)-2x-4y+1=0`
B. `x^(2)-y^(2)+2x-6y+2=0`
C. `x^(2)-y^(2)+4x-6y+5=0`
D. none of these

Explanation

To find the equation of the circle that touches the x-axis and has its center at (1, 2), we can follow these steps:

### Step 1: Identify the center and radius of the circle
The center of the circle is given as (1, 2). Since the circle touches the x-axis, the distance from the center to the x-axis is equal to the radius of the circle. The y-coordinate of the center is 2, which means the radius \( r \) is 2.

### Step 2: Write the standard form of the circle's equation
The standard form of the equation of a circle with center (h, k) and radius r is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Here, \( h = 1 \), \( k = 2 \), and \( r = 2 \).

### Step 3: Substitute the values into the equation
Substituting the values into the standard form, we have:
\[ (x - 1)^2 + (y - 2)^2 = 2^2 \]
This simplifies to:
\[ (x - 1)^2 + (y - 2)^2 = 4 \]

### Step 4: Expand the equation
Now, we will expand the equation:
\[ (x - 1)^2 = x^2 - 2x + 1 \]
\[ (y - 2)^2 = y^2 - 4y + 4 \]
Combining these, we get:
\[ x^2 - 2x + 1 + y^2 - 4y + 4 = 4 \]

### Step 5: Simplify the equation
Now, we simplify the equation:
\[ x^2 + y^2 - 2x - 4y + 1 + 4 - 4 = 0 \]
This simplifies to:
\[ x^2 + y^2 - 2x - 4y + 1 = 0 \]

### Final Answer
Thus, the equation of the circle is:
\[ x^2 + y^2 - 2x - 4y + 1 = 0 \]

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