Find the equation of a circle which touches the X-axis and whose centre is (2,3).

Options

A. `(x-2)^(2)+(y-3)^(2)=3^(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 (2, 3), we can follow these steps:

### Step 1: Identify the center and radius of the circle.
The center of the circle is given as (h, k) = (2, 3). Since the circle touches the X-axis, the distance from the center to the X-axis will be equal to the radius (r) of the circle.

### Step 2: Determine the radius.
The Y-coordinate of the center is 3, which means the radius of the circle is also 3. This is because the circle touches the X-axis at the point directly below the center, which is at (2, 0).

### Step 3: Use the standard equation of a circle.
The standard 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 into the equation:
- h = 2
- k = 3
- r = 3

The equation becomes:
\[ (x - 2)^2 + (y - 3)^2 = 3^2 \]

### Step 4: Simplify the equation.
Calculating \(3^2\):
\[ 3^2 = 9 \]
Thus, the equation simplifies to:
\[ (x - 2)^2 + (y - 3)^2 = 9 \]

### Final Equation:
The equation of the circle is:
\[ (x - 2)^2 + (y - 3)^2 = 9 \]

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