Question

Chords of the circle $x^2+y^2+2gx+2fy+c=0$ subtends a right angle at the origin. The locus of the feet of the perpendiculars from the origin to these chords is

• A. $x^2+y^2+gx+fy+c=0$
• B. $2(x^2+y^2)+gx+fy+c=0$
• C. $2(x^2+y^2+gx+fy)+c=0$
• D. $x^2+y^2+2(gx+fy+c)=0$

Answer 1

The correct option is C $2(x^2+y^2+gx+fy)+c=0$

Let $P(h,k)$ be the foot of perpendicular drawn from the origin to the chord $AB$ as shown such that $AB$ subtends a right angle at $O$.

Now, the equation of $AB$ is $y-k=-\frac{h}{k}(x-h)⟹hx+ky=h^2+k^2$.

The homogeneous equation of $OA$ and $OB$ can be found by homogenising each of terms of equation of the circle as

$x^2+y^2+(2gx+2fy)\left(\frac{hx+ky}{h^2+k^2}\right)+c{\left(\frac{hx+ky}{h^2+k^2}\right)}^2=0$

Now, as $OA$ and $OB$ subtend a right angle, coefficient of $x^2$ + coefficient of $y^2$ = 0

Hence, we have

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

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

$\therefore 2(h^2+k^2{)}^2+(2gh+2fk\left)\right(h^2+k^2)+c(h^2+k^2)=0$

$\therefore 2(h^2+k^2)+2gh+2fk+c=0$

Hence, the required answer is $2(x^2+y^2)+2gx+2fy+c=0$.