A circle touching the X-axis at (3, 0) and making a intercept of length 8 on the Y-axis passes through the point

Options

A. (3, 10)
B. (3, 5)
C. (2, 3)
D. (1, 5)

Explanation

To solve the problem of finding the equation of a circle that touches the x-axis at the point (3, 0) and makes an intercept of length 8 on the y-axis, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Center of the Circle**:
- Since the circle touches the x-axis at (3, 0), the x-coordinate of the center of the circle is 3. Let's denote the center as (3, h), where h is the y-coordinate of the center.
- The distance from the center to the x-axis (which is the radius) must be equal to h.

2. **Determine the Length of the Intercept**:
- The circle makes an intercept of length 8 on the y-axis. This means that the circle intersects the y-axis at two points, which are 4 units above and 4 units below the center (since the total length of the intercept is 8).
- Therefore, the y-coordinates of these points are h + 4 and h - 4.

3. **Setting Up the Equation**:
- Since the circle touches the x-axis, the radius (h) must be equal to 4 (the distance from the center to the points on the y-axis).
- Hence, we have h = 4.

4. **Finding the Center**:
- Now, substituting h back, the center of the circle is (3, 4).

5. **Calculating the Radius**:
- The radius of the circle is equal to the distance from the center to the x-axis, which is 4.

6. **Formulating the Equation of the 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 h = 3, k = 4, and r = 4 into the equation, we get:
\[ (x - 3)^2 + (y - 4)^2 = 4^2 \]
- Simplifying this gives:
\[ (x - 3)^2 + (y - 4)^2 = 16 \]

### Final Answer:
The equation of the circle is:
\[ (x - 3)^2 + (y - 4)^2 = 16 \]