If `g^(2)+f^(2)=c`, then the equation `x^(2)+y^(2)+2gx+2fy+c=0` will represent

Options

A. a circle of radius g
B. a circle of radius f
C. a circle of diameter `sqrt(c)`
D. a circle of radius 0

Explanation

To solve the problem, we start with the equation given:

1. **Given Equation**:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \( c = g^2 + f^2 \).

2. **Substituting for \( c \)**:
We substitute \( c \) in the equation:
\[ x^2 + y^2 + 2gx + 2fy + g^2 + f^2 = 0 \]

3. **Rearranging the Equation**:
We can rearrange the equation as follows:
\[ x^2 + 2gx + g^2 + y^2 + 2fy + f^2 = 0 \]

4. **Completing the Square**:
Now we can complete the square for both \( x \) and \( y \):
\[ (x + g)^2 + (y + f)^2 = 0 \]

5. **Analyzing the Equation**:
The equation \( (x + g)^2 + (y + f)^2 = 0 \) implies that the sum of two squares is equal to zero. This can only happen if both squares are individually zero:
\[ (x + g)^2 = 0 \quad \text{and} \quad (y + f)^2 = 0 \]

6. **Finding the Center**:
From the above, we find:
\[ x + g = 0 \quad \Rightarrow \quad x = -g \]
\[ y + f = 0 \quad \Rightarrow \quad y = -f \]
Thus, the center of the circle is at the point \( (-g, -f) \).

7. **Finding the Radius**:
Since the equation represents a point (the only solution is when both squares are zero), the radius is:
\[ r = 0 \]

8. **Conclusion**:
Therefore, the equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \) represents a **point circle** (or degenerate circle) with center at \( (-g, -f) \) and radius \( 0 \).