Question

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

• A. $x^2+y^2-hx-ky=0$
• B. $x^2+y^2+hx+ky=0$
• C. ${3x}^2+{3y}^2+hx-ky=0$
• D. $Noneofthese$

Answer 1

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

let $P(h,k)$ be the given point, let $Q(x,y)$ be the foot of the perpendicular, and Let $O$ be the origin. The line can have any direction.

$\angle PQO={90}^o$

point $Q$ lies on the circle having diameter $OP$.

The lower point $Q$:

$(x-0)(x-h)+(y-0)(y-k)=0$

$x^2+y^2-hx-ky=0$