Question

Let $S \equiv x^{2}+y^{2}+2 g x+2$ 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. $x^{2}+y^{2}-g x+f y+\frac{c}{2}=0$
• B. $x^{2}+y^{2}+g x+f y+\frac{c}{2}=0$
• C. $x^{2}+y^{2}+g x-f y+\frac{c}{2}-0$
• 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 $A B$. Equation of the chord $A B$ 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
$x^{2}+y^{2}+2 g x+2 f y + c=0$ ... (ii)
and the variable chord $AB$ is
$x ^{2}+ y ^{2}+2( gx + fy )\left(\frac{ hx + ky }{ h ^{2}+ k ^{2}}\right)+ c \left(\frac{ h x + ky }{ h ^{2}+ k ^{2}}\right)^{2}=0$ ...(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., $\left(1+\frac{2 gh }{ h ^{2}+ k ^{2}}+\frac{ ch ^{2}}{\left( h ^{2}+ k ^{2}\right)^{2}}\right)+\left(1+\frac{2 fk }{ h ^{2}+ k ^{2}}+\frac{ ck ^{2}}{\left( h ^{2}+ k ^{2}\right)^{2}}\right)=0$
Putting $(x, y)$ in place of $(h, k)$ gives the equation of the required locus as
$x^{2}+y^{2}+g x+f y+\frac{c}{2}=0$