If O is the origin and OP, OQ are tangents to the circle `x^(2)+y^(2)+2g x+2fy+c=0 (c ne 0)`, then the equation of the circumcircle of the triangle OPQ is given by

Options

A. `x^(2)+y^(2)+2g x +2fy =0`
B. `x^(2)+y^(2) -g x-fy=0`
C. `x^(2)+y^(2) +g x +fy=0`
D. none of these

Answer

To solve the problem of finding the equation of the circumcircle of triangle OPQ, where O is the origin and OP, OQ are tangents to the circle given by the equation \(x^2 + y^2 + 2gx + 2fy + c = 0\) (with \(c \neq 0\)), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Circle and Tangents**:
The given circle is \(S: x^2 + y^2 + 2gx + 2fy + c = 0\). The point O (0, 0) is the origin from which the tangents OP and OQ are drawn.

2. **Find the Chord of Contact**:
The chord of contact from the point O (0, 0) to the circle is given by the equation:
\[ Gx + Fy + C = 0 \]
Here, substituting \(x_1 = 0\) and \(y_1 = 0\) in the general form of the chord of contact gives us:
\[ G(0) + F(0) + c = 0 \implies Gx + Fy + c = 0 \]
Thus, the equation of the chord of contact (line L) is:
\[ Gx + Fy + c = 0 \tag{1} \]

3. **Set Up the Equation of the Circumcircle**:
The circumcircle of triangle OPQ will pass through the points O, P, and Q. The equation of the circumcircle can be expressed as:
\[ S + \lambda L = 0 \]
where \(S\) is the equation of the given circle and \(L\) is the equation of the chord of contact.

4. **Substituting the Values**:
Substitute the equations:
\[ x^2 + y^2 + 2gx + 2fy + c + \lambda (Gx + Fy + c) = 0 \]
This simplifies to:
\[ x^2 + y^2 + (2g + \lambda G)x + (2f + \lambda F)y + (c + \lambda c) = 0 \]

5. **Condition for Passing through the Origin**:
Since the circumcircle must pass through the origin (0, 0), we substitute \(x = 0\) and \(y = 0\):
\[ c + \lambda c = 0 \implies c(1 + \lambda) = 0 \]
Since \(c \neq 0\), we have:
\[ 1 + \lambda = 0 \implies \lambda = -1 \]

6. **Final Equation of the Circumcircle**:
Substitute \(\lambda = -1\) back into the equation:
\[ x^2 + y^2 + (2g - G)x + (2f - F)y + (c - c) = 0 \]
This simplifies to:
\[ x^2 + y^2 + (2g - G)x + (2f - F)y = 0 \]

7. **Substituting G and F**:
Recall that \(G = g\) and \(F = f\), so we have:
\[ x^2 + y^2 + gx + fy = 0 \]

### Conclusion:
Thus, the equation of the circumcircle of triangle OPQ is:
\[ \boxed{x^2 + y^2 + gx + fy = 0} \]