Question

Let $S\equiv x^2+y^2+2gx+2$ fy $+c=0$ be a given circle. The locus of the foot of the perpendicular drawn from the origin upon any chord of $S$ which subtends right angle at the origin, is.

• A. $x^2+y^2-gx+fy+\frac{c}{2}=0$
• B. $x^2+y^2+gx+fy+\frac{c}{2}=0$
• C. $x^2+y^2+gx-fy+\frac{c}{2}-0$
• D. None of these

Answer 1

Answer: (B)

$AB$ is a variable chord such that $=\angle AOB=\frac{\pi }{2}$.
Let $P(h,k)$ be the foot of the perpendicular drawn from origin upon $AB$. Equation of the chord $AB$ is
$y-k=\frac{-h}{k}(x-h)$
i.e., $hx+ky=h^2+k^2$ ....(i)
Equation of the pair of straight lines passing through the origin and the intersection point of the given circle
$x^2+y^2+2gx+2fy+c=0$ ... (ii)
and the variable chord $AB$ is
$x^2+y^2+2(gx+fy)\left(\frac{hx+ky}{h^2+k^2}\right)+c{\left(\frac{hx+ky}{h^2+k^2}\right)}^2=0$ ...(iii)
If equation (iii) must represent a pair of perpendicular lines, then we have coeff. of $x^2+$ coeff. of $y^2=0$
i.e., $\left(1+\frac{2gh}{h^2+k^2}+\frac{c{h}^2}{{\left(h^2+k^2\right)}^2}\right)+\left(1+\frac{2fk}{h^2+k^2}+\frac{c{k}^2}{{\left(h^2+k^2\right)}^2}\right)=0$
Putting $(x,y)$ in place of $(h,k)$ gives the equation of the required locus as
$x^2+y^2+gx+fy+\frac{c}{2}=0$