Suppose a point `(x_(1),y_(1))` satisfies `x^(2)+y^(2)+2gx+2fy+c=0` then show that it represents a circle whenever g,f and c are real.

Answer

To show that the equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \) represents a circle whenever \( g, f, \) and \( c \) are real, we can follow these steps:

### Step 1: Identify the General Form of the Equation
The given equation is of the form:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
This is a second-degree equation in \( x \) and \( y \).

### Step 2: Recognize the Condition for a Circle
A conic section represents a circle if the coefficient of \( xy \) (which is \( h \)) is zero, and the coefficients of \( x^2 \) and \( y^2 \) are equal. In our case, we have:
- Coefficient of \( x^2 \) is \( 1 \)
- Coefficient of \( y^2 \) is \( 1 \)
- Coefficient of \( xy \) is \( 0 \) (since there is no \( xy \) term)

Since \( h = 0 \), the equation can represent a circle.

### Step 3: Rearranging the Equation
We can rearrange the equation to express it in standard form for a circle. We start by completing the square for both \( x \) and \( y \):
\[ x^2 + 2gx + y^2 + 2fy + c = 0 \]
Completing the square for \( x \):
\[ (x + g)^2 - g^2 \]
Completing the square for \( y \):
\[ (y + f)^2 - f^2 \]
Substituting these back into the equation gives:
\[ (x + g)^2 - g^2 + (y + f)^2 - f^2 + c = 0 \]
This simplifies to:
\[ (x + g)^2 + (y + f)^2 = g^2 + f^2 - c \]

### Step 4: Analyzing the Right Side
For this equation to represent a circle, the right side must be positive:
\[ g^2 + f^2 - c > 0 \]
This condition ensures that the radius of the circle is real and positive.

### Conclusion
Since \( g, f, \) and \( c \) are real numbers, \( g^2 \) and \( f^2 \) are non-negative. Therefore, if \( g^2 + f^2 > c \), it confirms that the equation represents a circle.

Thus, we conclude that the equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \) represents a circle whenever \( g, f, \) and \( c \) are real.

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