A variable line cuts the axes of co ordinates in points A and B such that `OA+OB=c`. The locus of foot of perpendicular from origin to the line is

Options

A. `x^(2)+y^(2)=cxy`
B. `x^(2)+y^(2)=2cxy`
C. `(x+y)(x^(2)+y^(2))=cxy`
D. None of these

Answer

To solve the problem, we need to find the locus of the foot of the perpendicular from the origin to a variable line that cuts the coordinate axes at points A and B such that \( OA + OB = c \).

### Step-by-Step Solution:

1. **Identify Points A and B**:
- Let the line cut the x-axis at point A and the y-axis at point B.
- The coordinates of point A can be represented as \( (a, 0) \) and the coordinates of point B as \( (0, b) \).

2. **Express the Condition**:
- According to the problem, the condition given is \( OA + OB = c \).
- The distances OA and OB can be expressed as:
\[ OA = a \quad \text{and} \quad OB = b \]
- Therefore, the condition becomes:
\[ a + b = c \]

3. **Equation of the Line**:
- The equation of the line passing through points A and B can be written in intercept form:
\[ \frac{x}{a} + \frac{y}{b} = 1 \]

4. **Finding the Foot of the Perpendicular**:
- The foot of the perpendicular from the origin (0, 0) to the line can be denoted as point \( (h, k) \).
- The slope of the line AB is given by:
\[ \text{slope of AB} = -\frac{b}{a} \]
- The slope of the line from the origin to point M (the foot of the perpendicular) is:
\[ \text{slope of OM} = \frac{k}{h} \]
- Since these two lines are perpendicular, we have:
\[ -\frac{b}{a} \cdot \frac{k}{h} = -1 \]
- This simplifies to:
\[ \frac{b}{a} = \frac{h}{k} \quad \Rightarrow \quad bh = ak \]

5. **Substituting for a and b**:
- From the condition \( a + b = c \), we can express \( b \) in terms of \( a \):
\[ b = c - a \]
- Substitute this into the equation \( bh = ak \):
\[ (c - a)h = ak \]
- Rearranging gives:
\[ ch - ah = ak \quad \Rightarrow \quad ch = ak + ah \quad \Rightarrow \quad ch = a(k + h) \]
- Thus, we can express \( a \):
\[ a = \frac{ch}{k + h} \]

6. **Finding b**:
- Substitute \( a \) back into the equation for \( b \):
\[ b = c - a = c - \frac{ch}{k + h} = \frac{c(k + h) - ch}{k + h} = \frac{ck}{k + h} \]

7. **Final Locus Equation**:
- Now substitute \( a \) and \( b \) back into the line equation:
\[ \frac{x}{\frac{ch}{k + h}} + \frac{y}{\frac{ck}{k + h}} = 1 \]
- This simplifies to:
\[ \frac{(k + h)x}{ch} + \frac{(k + h)y}{ck} = 1 \]
- Multiplying through by \( ckh \) gives:
\[ kx + hy = cxy \]
- Rearranging leads to:
\[ (x + y)(x^2 + y^2) = cxy \]

### Final Answer:
The locus of the foot of the perpendicular from the origin to the line is given by:
\[ (x + y)(x^2 + y^2) = cxy \]