For a variabkle line `x/a+y/b=1` where `1/(a^(2))+1/(b^(2))=1/(c^(2))` the locus of the foot of perpendicular drawn from origin to it is

Options

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

Answer

To find the locus of the foot of the perpendicular drawn from the origin to the variable line given by the equation \( \frac{x}{a} + \frac{y}{b} = 1 \) under the condition \( \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2} \), we can follow these steps:

### Step 1: Rewrite the line equation
The line can be rewritten in standard form:
\[ \frac{x}{a} + \frac{y}{b} = 1 \implies bx + ay = ab \]
This represents a line in the Cartesian plane.

### Step 2: Find the distance from the origin to the line
The formula for the distance \( d \) from a point \( (x_0, y_0) \) to a line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line \( bx + ay - ab = 0 \), we have \( A = b \), \( B = a \), and \( C = -ab \). The distance from the origin \( (0, 0) \) to the line is:
\[ d = \frac{|b(0) + a(0) - ab|}{\sqrt{b^2 + a^2}} = \frac{ab}{\sqrt{a^2 + b^2}} \]

### Step 3: Use the given condition
According to the problem, we have:
\[ \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2} \]
This can be rearranged to give:
\[ \frac{a^2 + b^2}{a^2 b^2} = \frac{1}{c^2} \implies c^2(a^2 + b^2) = a^2 b^2 \]

### Step 4: Set the distance equal to \( c \)
From the distance formula, we have:
\[ \frac{ab}{\sqrt{a^2 + b^2}} = c \]
Squaring both sides gives:
\[ \frac{a^2 b^2}{a^2 + b^2} = c^2 \]

### Step 5: Substitute \( c^2 \)
Using the earlier result \( c^2(a^2 + b^2) = a^2 b^2 \), we can substitute \( c^2 \) into the equation:
\[ c^2 = \frac{a^2 b^2}{a^2 + b^2} \]

### Step 6: Express in terms of \( x \) and \( y \)
Now, let \( h = x \) and \( k = y \) (the coordinates of the foot of the perpendicular). We have:
\[ x^2 + y^2 = c^2 \]
Substituting for \( c^2 \):
\[ x^2 + y^2 = \frac{a^2 b^2}{a^2 + b^2} \]

### Step 7: Conclusion
The locus of the foot of the perpendicular from the origin to the line is a circle centered at the origin with radius \( c \):
\[ x^2 + y^2 = c^2 \]