Circle(s) touching x-axis at a distance 3 from the origin and having an intercept of length `2sqrt7` on y-axis is (are)

Options

A. `x^(2)+y^(2)-6x+8y+9=0`
B. `x^(2)+y^(2)-6x+7y+9=0`
C. `x^(2)+y^(2)-6x-8y+9=0`
D. `x^(2)=y^(2)-6x-7y+9=0`

Explanation

To find the equations of the circles that touch the x-axis at a distance of 3 from the origin and have an intercept of length \(2\sqrt{7}\) 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 a distance of 3 from the origin, the center of the circle will be at the point \((3, r)\) where \(r\) is the radius of the circle. The y-coordinate of the center is equal to the radius because the circle touches the x-axis.

2. **Determine the radius**:
The circle has an intercept of length \(2\sqrt{7}\) on the y-axis. This means that the distance from the center of the circle to the y-axis is equal to the radius \(r\). The intercept on the y-axis can be represented as the vertical distance from the center to the points where the circle intersects the y-axis.

The points of intersection on the y-axis can be calculated as:
\[ y = r + d \quad \text{and} \quad y = r - d \]
where \(d\) is half the length of the intercept. Since the total intercept is \(2\sqrt{7}\), we have:
\[ d = \sqrt{7} \]
Therefore, the points of intersection are:
\[ y = r + \sqrt{7} \quad \text{and} \quad y = r - \sqrt{7} \]

3. **Set up the equation for the radius**:
The distance between these two points is given by:
\[ (r + \sqrt{7}) - (r - \sqrt{7}) = 2\sqrt{7} \]
This confirms that the distance between the two points on the y-axis is indeed \(2\sqrt{7}\).

4. **Establish the relationship**:
We know that the radius \(r\) must equal the distance from the center to the x-axis, which is 3. Therefore, we have:
\[ r = 3 \]

5. **Find the coordinates of the center**:
The center of the circle is at \((3, 3)\) since \(r = 3\).

6. **Write the equation of the circle**:
The standard form of the equation of a circle with center \((h, k)\) and radius \(r\) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \(h = 3\), \(k = 3\), and \(r = 3\):
\[ (x - 3)^2 + (y - 3)^2 = 3^2 \]
Simplifying gives:
\[ (x - 3)^2 + (y - 3)^2 = 9 \]

7. **Consider other quadrants**:
Since the circle could also be in other quadrants, we can also consider the center at \((3, -3)\) or \((-3, 3)\) and \((-3, -3)\) as potential centers. Thus, the equations of the circles can also be:
\[ (x - 3)^2 + (y + 3)^2 = 9 \]
\[ (x + 3)^2 + (y - 3)^2 = 9 \]
\[ (x + 3)^2 + (y + 3)^2 = 9 \]

### Final Equations of the Circles:
1. \((x - 3)^2 + (y - 3)^2 = 9\)
2. \((x - 3)^2 + (y + 3)^2 = 9\)
3. \((x + 3)^2 + (y - 3)^2 = 9\)
4. \((x + 3)^2 + (y + 3)^2 = 9\)