The circle with centre (4,-1) and touching x-axis is

Options

A. `x^(2)+y^(2)-8x+2y+16=0`
B. `x+y^(2)+18x-2y-16=0`
C. `x^(2)+y^(2)-4x+y+4=0`
D. `x^(2)+y^(2)+14x-y+4=0`

Answer

To find the equation of the circle with center at (4, -1) that touches the x-axis, 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) = (4, -1). Since the circle touches the x-axis, the distance from the center to the x-axis is equal to the radius. The y-coordinate of the center is -1, which means the distance to the x-axis (y = 0) is:
\[ \text{Distance} = |k| = |-1| = 1 \]
Thus, the radius \( r \) of the circle is 1.

### Step 2: Use the center-radius form of the circle's equation
The equation of a circle in center-radius form is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting the values of h, k, and r:
- \( h = 4 \)
- \( k = -1 \)
- \( r = 1 \)

We get:
\[ (x - 4)^2 + (y + 1)^2 = 1^2 \]
This simplifies to:
\[ (x - 4)^2 + (y + 1)^2 = 1 \]

### Step 3: Expand the equation
Now we will expand the equation:
1. Expand \( (x - 4)^2 \):
\[ (x - 4)^2 = x^2 - 8x + 16 \]
2. Expand \( (y + 1)^2 \):
\[ (y + 1)^2 = y^2 + 2y + 1 \]

Putting it all together:
\[ x^2 - 8x + 16 + y^2 + 2y + 1 = 1 \]

### Step 4: Simplify the equation
Combine like terms:
\[ x^2 + y^2 - 8x + 2y + 16 + 1 - 1 = 0 \]
This simplifies to:
\[ x^2 + y^2 - 8x + 2y + 16 = 0 \]

### Conclusion
The equation of the circle is:
\[ x^2 + y^2 - 8x + 2y + 16 = 0 \]

### Final Answer
Thus, the correct option is the first one:
\[ x^2 + y^2 - 8x + 2y + 16 = 0 \]

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