Find the equation of a circle which touches the `X-axis` and whose centre is `(1,2)`.

Options

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

Explanation

To find the equation of a circle that touches the X-axis and has its center at the point (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). The Y-coordinate of the center (2) indicates that the circle is located 2 units above the X-axis.

### Step 2: Determine the radius of the circle
Since the circle touches the X-axis, the radius of the circle is equal to the Y-coordinate of the center. Therefore, the radius \( r \) is:
\[ r = 2 \]

### Step 3: Write the standard equation of the circle
The standard form of the equation of a circle with center \((h, k)\) and radius \(r\) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting the values of the center (1, 2) and the radius (2) into the equation, we have:
\[ (x - 1)^2 + (y - 2)^2 = 2^2 \]

### Step 4: Simplify the equation
Calculating \(2^2\) gives us \(4\). Thus, the equation becomes:
\[ (x - 1)^2 + (y - 2)^2 = 4 \]

### Final Answer
The equation of the circle is:
\[ (x - 1)^2 + (y - 2)^2 = 4 \]

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