Question

A straight line passes through a fixed point $(h,k)$. The locus of the foot of perpendicular on it drawn from the origin is

• A. $x^2+y^2=hx-ky$
• B. $x^2+y^2=kx-hy$
• C. $x^2+y^2-hx-ky=0$
• D. $x^2+y^2-kx-hy=0$

Answer 1

The correct option is C $x^2+y^2-hx-ky=0$

Let the line passing through $(h,k)$ has slope $m$ then equation of the line will be
$y-k=m(x-h)\cdots \left(1\right)$
Now the equation of the perpendicular drawn from origin to the given line will be
$x=-my$
so putting the value of $m$ in $\left(1\right)$
$x^2+y^2-hx-ky=0$ as the locus

Alternate solution:
Let the locus of the foot of perpendicualar be $(a,b)$.
Let the line passing through $(h,k)$ has slope $m$ then equation of the line will be
$y-k=m(x-h)\Rightarrow mx-y-mh+k=0$
using foot of perpendicular formulae
$\frac{a-0}{m}=\frac{b-0}{-1}=-\frac{(-mh+k)}{m^2+1}$
$\Rightarrow m=-\frac{a}{b},$
$b=\frac{k-mh}{1+m^2}$
eliminating the $m$ from above equations
$a^2+b^2-ha-kb=0$
$\therefore$ locus will be
$x^2+y^2-hx-ky=0$