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. $x^2+y^2-hx-ky=0$
• B. $x^2+y^2+hx+ky=0$
• C. $3x^2+3y^2+hx-ky=0$
• D. None of these

Answer 1

Answer: (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 diretcion

$\angle PQO={90}^{\circ }$

Point Q lies on the circle having diameter OP.

The locus point Q

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

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