Question

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

• A. $x^2+y^2+gx+fy+c/2=0$
• B. $x^2+y^2=g$
• C. $x^2+y^2=f$
• D. $x^2+y^2+g=0$

Answer 1

Answer: (A)

$x^2+y^2+gx+fy+c/2=0$

Given circle $=x^2+y^2+2gx+2fy+c=0$

Let $LM$ be the chord

of circle $C$ and subtend right angle at $(0,0)O,P(h,k)$ be fool of $\perp$ from $O$ on chord $LM$

as $OP\perp ML$, eqn of $LM$ is:

$y-k=\frac{h}{k}(x-h)$

$hx+ky=h^2+k^2$ .......... $\left(1\right)$

equation of pair of line joining point of intersection of $\left(1\right)$ with $S=0\Rightarrow x^2+y^2+2(gx+fy)\frac{(hx+ky}{h^2+k^2}+c\frac{(hx+ky{)}^2}{(h^2+k^2{)}^2}=0$ ........... $\left(2\right)$

as lines in $\left(2\right)$ are at right angles

$1+1=\frac{2gh+2fk}{h^2+k^2}+c\frac{(h^2+k^2)}{(h^2+k^2)}=0$

$h^2+k^2+gh+fk+\frac{c}{2}=0$

$\therefore$ Locus of $(h,k)$ is $\Rightarrow S=x^2+y^2+gx+fy+\frac{c}{2}=0$

Option $\left(A\right)$