Equation of circle having radius 5, and touching X-axis at (-1,0), is

Options

A. `x^(2) + y^(2) pm 2x + 10y - 1 = 0 `
B. `x^(2) + y^(2) pm 2x - 10y + 1= 0 `
C. `x^(2) + y^(2) + 2x + 10y pm 1 = 0 `
D. `x^(2) + y^(2) + 2x pm 10y + 1 = 0`

Answer

To find the equation of a circle with a radius of 5 that touches the X-axis at the point (-1, 0), we can follow these steps:

### Step 1: Identify the center of the circle
Since the circle touches the X-axis at (-1, 0), the x-coordinate of the center of the circle must be -1. The distance from the center to the X-axis is equal to the radius of the circle, which is 5. Therefore, the y-coordinate of the center can be either 5 (above the X-axis) or -5 (below the X-axis).

### Step 2: Determine the possible centers
The two possible centers of the circle are:
1. Center at (-1, 5)
2. Center at (-1, -5)

### Step 3: Write the equation of the circle
The general equation of a circle with center (h, k) and radius r is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
For the center (-1, 5) and radius 5:
\[ (x + 1)^2 + (y - 5)^2 = 5^2 \]
This simplifies to:
\[ (x + 1)^2 + (y - 5)^2 = 25 \]

For the center (-1, -5) and radius 5:
\[ (x + 1)^2 + (y + 5)^2 = 5^2 \]
This simplifies to:
\[ (x + 1)^2 + (y + 5)^2 = 25 \]

### Step 4: Final equations
The two equations of the circles are:
1. \((x + 1)^2 + (y - 5)^2 = 25\) (Circle above the X-axis)
2. \((x + 1)^2 + (y + 5)^2 = 25\) (Circle below the X-axis)