The equation of the tangent to circle `x^(2)+y^(2)+2g x+2fy=0` at origin is :

Options

A. `fx+gy=0`
B. `gx+fy=0`
C. `x=0`
D. `y=0`

Answer

To find the equation of the tangent to the circle given by the equation \( x^2 + y^2 + 2gx + 2fy = 0 \) at the origin (0, 0), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Circle Equation**: The given equation of the circle is:
\[ x^2 + y^2 + 2gx + 2fy = 0 \]

2. **Point of Tangency**: We need to find the equation of the tangent at the point (0, 0).

3. **Substituting for Tangent**: The general formula for the equation of a tangent to the circle at a point \((x_1, y_1)\) is:
\[ xx_1 + yy_1 + g(x + x_1) + f(y + y_1) = 0 \]
Here, \(x_1 = 0\) and \(y_1 = 0\).

4. **Plugging in the Values**: Substitute \(x_1\) and \(y_1\) into the tangent equation:
\[ x(0) + y(0) + g(x + 0) + f(y + 0) = 0 \]
This simplifies to:
\[ gx + fy = 0 \]

5. **Final Equation of the Tangent**: Thus, the equation of the tangent to the circle at the origin is:
\[ gx + fy = 0 \]

### Final Answer:
The equation of the tangent to the circle \(x^2 + y^2 + 2gx + 2fy = 0\) at the origin is:
\[ gx + fy = 0 \]