Question

If the line $\dfrac{x}{a} + \dfrac{y}{b} = 1$ moves in such way that $\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} = \dfrac{1}{{{c^2}}}$​, where c is a constant, then the locus of the foot of the perpendicular from the origin on the straight line described the circle
A. ${x^2} + {y^2} = 4{c^2}$
B. ${x^2} + {y^2} = 2{c^2}$
C. ${x^2} + {y^2} = {c^2}$
D. None of these

Answer 1

Hint: The equation of a line is given. We can find the equation of the line perpendicular to the given line by using slope of the given line and the relationship between slopes of perpendicular lines. On squaring and adding the two equations we can obtain the locus of the foot of the perpendicular.

Complete step by step solution:
The equation of the line given is $\dfrac{x}{a} + \dfrac{y}{b} = 1...(1)$
The equation of line passing through the origin and perpendicular to line (1) is $\dfrac{x}{b} - \dfrac{y}{a} = 0{\text{ }}...\left( 2 \right)$
The foot of the perpendicular from origin on the line (1) is the point of intersection of lines (1) and (2)
So, in order to find its locus, we have to eliminate the variables a and b
Squaring and adding equations (1) and (2) we obtain,
${\left( {\dfrac{x}{a} + \dfrac{y}{b}} \right)^2} + {\left( {\dfrac{x}{b} - \dfrac{y}{a}} \right)^2} = 1 + 0$
$ \Rightarrow {x^2}\left( {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}} \right) + {y^2}\left( {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}} \right) = 1$
$ \Rightarrow \dfrac{{{x^2}}}{{{c^2}}} + \dfrac{{{y^2}}}{{{c^2}}} = 1$ $\left[ {\because \dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} = \dfrac{1}{{{c^2}}}} \right]$
$ \Rightarrow {x^2} + {y^2} = {c^2}$

Option ‘C’ is correct

Note: In order to solve such kinds of questions one must know how to find the equation of line when a point and a perpendicular line is given. This can be done by finding the slope of the line using the appropriate formula.