Question

The equation of the locus of the foot of perpendiculars drawn from the origin to the line passing through a fixed point $(a,b)$is

• A. $x^2+y^2–ax–by=0$
• B. $x^2+y^2+ax+by=0$
• C. $x^2+y^2–2ax–2by=0$
• D. None of these

Answer 1

The correct option is A

$x^2+y^2–ax–by=0$

Explanation of the correct option.

Step$1$: Find the slope of the line passing through the given point.

We know that equation of line passing through $(x_1,y_1)$ is given by

$(y–y_1)=m(x–x_1)$

Therefore, the equation of line passing through $(a,b)$is,

$(y–b)=m(x–a)$…………..$\left(1\right)$

Since the slope of the perpendicular line is $-\frac{1}{m}$.

Therefore, the equation of perpendicular line passing through the origin is,

$(y–0)=-\frac{1}{m}(x–0)$

$\Rightarrow$ $y=-\frac{x}{m}$

$\Rightarrow$ $m=-\frac{x}{y}$……………….$\left(2\right)$

Step$2$: Find the equation of the locus.

The foot of the perpendicular is the intersection of $\left(1\right)$ and $\left(2\right)$ and its locus is given by,

substitute the value of $m$ from equation $\left(2\right)$ to equation $\left(1\right)$.

$(y–b)=-\frac{x}{y}(x–a)$

$\Rightarrow$$y^2–by+x^2–ax=0$

$\Rightarrow$$x^2+y^2–ax–by=0$

Hence Option $\left(\mathrm{A}\right)$ is the correct answer.