The equation of the circle which touches x-axis and whose centre is (1,2) is

Options

A. (a) `x^(2) + y^(2)- 2x - 4y + 4 = 0 `
B. (b) `x^(2) + y^(2) - 2x - 4y + 1= 0 `
C. (c) ` x^(2) + y^(2) + 2x + 4y - 1 = 0 `
D. (d) `x^(2) + y^(2) + 2x - 4y + 1 = 0 `

Answer

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
The center of the circle is given as (h, k) = (1, 2). Since the circle touches the x-axis, the radius (r) of the circle is equal to the y-coordinate of the center, which is k = 2.

### Step 2: Write the equation of the circle
The general equation of a circle with center (h, k) and radius r is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting the values of h, k, and r:
\[ (x - 1)^2 + (y - 2)^2 = 2^2 \]
This simplifies to:
\[ (x - 1)^2 + (y - 2)^2 = 4 \]

### Step 3: Expand the equation
Now we will expand the left-hand side:
\[ (x - 1)^2 = x^2 - 2x + 1 \]
\[ (y - 2)^2 = y^2 - 4y + 4 \]
Combining these, we have:
\[ x^2 - 2x + 1 + y^2 - 4y + 4 = 4 \]
This simplifies to:
\[ x^2 + y^2 - 2x - 4y + 5 = 4 \]

### Step 4: Rearranging the equation
Now, we will move 4 to the left side:
\[ x^2 + y^2 - 2x - 4y + 5 - 4 = 0 \]
This simplifies to:
\[ x^2 + y^2 - 2x - 4y + 1 = 0 \]

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

### Final Answer
The correct option is:
**Option B: \( x^2 + y^2 - 2x - 4y + 1 = 0 \)**

---