Tangents `OP` and `OQ` are drawn from the origin o to the circle `x^2 + y^2 + 2gx + 2fy+c=0.` Find the equation of the circumcircle of the triangle `OPQ`.

Options

A. `((-g)/(2), (-f)/(2))`
B. `(g, f)`
C. `(-f, -g)`
D. `(f, g)`

Answer

To find the equation of the circumcircle of triangle \( OPQ \) where \( O \) is the origin and \( P \) and \( Q \) are points where tangents from the origin touch the circle given by the equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \), we can follow these steps:

### Step 1: Identify the center and radius of the circle
The given circle can be rewritten in standard form. The center \( C \) of the circle is given by the coordinates \( (-g, -f) \).

### Step 2: Understand the properties of the circumcircle
The circumcircle of triangle \( OPQ \) will pass through the points \( O \), \( P \), and \( Q \). A key property is that the circumcircle will also pass through the center of the original circle \( C \).

### Step 3: Determine the diameter of the circumcircle
Since \( OP \) and \( OQ \) are tangents from the origin to the circle, the angles \( \angle OCP \) and \( \angle OCQ \) are right angles (90 degrees). This means that the line segment \( OC \) is the diameter of the circumcircle.

### Step 4: Use the diameter to find the equation of the circumcircle
The endpoints of the diameter are \( O(0, 0) \) and \( C(-g, -f) \). The general equation of a circle with diameter endpoints \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:

\[ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \]

Substituting \( (x_1, y_1) = (0, 0) \) and \( (x_2, y_2) = (-g, -f) \):

\[ (x - 0)(x + g) + (y - 0)(y + f) = 0 \]

This simplifies to:

\[ x(x + g) + y(y + f) = 0 \]

### Step 5: Expand and rearrange the equation
Expanding the equation gives:

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

This is the required equation of the circumcircle of triangle \( OPQ \).

### Final Answer
The equation of the circumcircle of triangle \( OPQ \) is:

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