If the circle `x^(2) + y^(2) + 2g x + 8y + 16 = 0` touches the x axis, then the values of g are

Options

A. A. `pm 16`
B. B. ` pm 8`
C. C. ` pm 4 `
D. D. ` pm 2 `

Answer

To solve the problem, we need to determine the values of \( g \) for which the circle given by the equation

\[ x^2 + y^2 + 2gx + 8y + 16 = 0 \]

touches the x-axis.

### Step 1: Identify the standard form of the circle equation
The general form of a circle's equation is given by:

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

From the given equation, we can identify:
- \( 2g \) as the coefficient of \( x \)
- \( 2f \) as the coefficient of \( y \)
- \( c = 16 \)

Thus, we have:
- \( f = 4 \) (since \( 2f = 8 \))
- \( c = 16 \)

### Step 2: Use the condition for the circle to touch the x-axis
For a circle to touch the x-axis, the distance from the center of the circle to the x-axis must equal the radius of the circle. The center of the circle can be found from the coefficients \( g \) and \( f \):

\[ \text{Center} = (-g, -f) = (-g, -4) \]

The radius \( r \) of the circle is given by:

\[ r = \sqrt{g^2 + f^2 - c} \]

Substituting the values of \( f \) and \( c \):

\[ r = \sqrt{g^2 + 4^2 - 16} = \sqrt{g^2 + 16 - 16} = \sqrt{g^2} \]

Thus, the radius simplifies to:

\[ r = |g| \]

### Step 3: Set up the equation for touching the x-axis
The distance from the center to the x-axis is simply \( |-4| = 4 \). For the circle to touch the x-axis, we set the radius equal to this distance:

\[ |g| = 4 \]

### Step 4: Solve for \( g \)
This absolute value equation gives us two cases:

1. \( g = 4 \)
2. \( g = -4 \)

Thus, the values of \( g \) are:

\[ g = \pm 4 \]

### Conclusion
The values of \( g \) for which the circle touches the x-axis are \( g = 4 \) and \( g = -4 \).

### Final Answer
The values of \( g \) are \( \pm 4 \).

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