Find the locus of the foot of the perpendicular let fall from the origin upon any chord of the circle `x^(2) + y^(2) + 2gx + 2fy + c = 0 ` which subtends a right angle at the origin.

Answer

To find the locus of the foot of the perpendicular dropped from the origin onto any chord of the circle given by the equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \) that subtends a right angle at the origin, we can follow these steps:

### Step 1: Understanding the Circle and Chord
The given equation represents a circle with center at \( (-g, -f) \) and radius \( \sqrt{g^2 + f^2 - c} \). A chord of this circle subtending a right angle at the origin means that the angle between the lines connecting the origin to the endpoints of the chord is \( 90^\circ \).

### Step 2: Equation of the Chord
Let the endpoints of the chord be \( A(x_1, y_1) \) and \( B(x_2, y_2) \). The condition for the chord \( AB \) to subtend a right angle at the origin can be expressed using the slopes of the lines \( OA \) and \( OB \):
\[ \frac{y_1}{x_1} \cdot \frac{y_2}{x_2} = -1 \]
This implies that:
\[ y_1 y_2 + x_1 x_2 = 0 \]

### Step 3: Equation of the Chord
The equation of the chord \( AB \) can be derived using the two-point form of the line equation. The line passing through points \( A \) and \( B \) can be expressed as:
\[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \]
This can be rearranged to form the general line equation \( Ax + By + C = 0 \).

### Step 4: Finding the Foot of the Perpendicular
Let \( P(\alpha, \beta) \) be the foot of the perpendicular from the origin to the chord \( AB \). The slope of the line from the origin to point \( P \) is \( \frac{\beta}{\alpha} \). The slope of the line \( AB \) must be the negative reciprocal of this slope since they are perpendicular.

### Step 5: Homogenization
To find the locus of point \( P \), we can use the method of homogenization. We will replace the coordinates \( (x_1, y_1) \) and \( (x_2, y_2) \) with \( (\alpha, \beta) \) and express the condition of the chord in terms of \( \alpha \) and \( \beta \).

### Step 6: Setting Up the Homogenized Equation
The homogenized equation can be set as:
\[ x^2 + y^2 + 2g\alpha + 2f\beta + c = 0 \]
This represents the condition for the locus of the foot of the perpendicular.

### Step 7: Final Locus Equation
To derive the final locus equation, we will set:
\[ \alpha^2 + \beta^2 + g\alpha + f\beta + \frac{c}{2} = 0 \]
Replacing \( \alpha \) and \( \beta \) with \( x \) and \( y \), we get:
\[ x^2 + y^2 + gx + fy + \frac{c}{2} = 0 \]

### Conclusion
Thus, the locus of the foot of the perpendicular from the origin to any chord of the circle that subtends a right angle at the origin is given by:
\[ x^2 + y^2 + gx + fy + \frac{c}{2} = 0 \]