Question

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

• A. $2(x^2+y^2)+2(9x+fy)+c=0$
• B. $x^2+y^2+2(9x+fy)+c=0$
• C. $2x^2+2y^2+9xfy+c=0$
• D. None of these

Answer 1

The correct option is D None of these

Let, $P(h,k)$ be the foot of perpendicular drawn from origin $O(0,0)$ on the chord AB of the given circle such that the chord AB subtends a right angle at the origin.
The equation of chord AB is
$y-k=-\frac{h}{k}(x-h)\Rightarrow hx+ky=h^2+k^2$
The combined equation of OA and OB is homogeneous equation of the second degree obtained by the help of the given circle and the chord AB and is given by
$x^2+y^2(2gx+2fy)\left(\frac{hx+ky}{h^2+k^2}\right)+c{\left(\frac{hx+ky}{h^2+k^2}\right)}^2=0$
Since, the lines OA and OB are at right angle.
Therefore coefficient of $x^2+$ coefficient of $y^2=0$
$\Rightarrow \{1+\frac{2gh}{h^2+k^2}+\frac{{ch}^2}{{(h^2+k^2)}^2}\}+\{1+\frac{2fk}{h^2+k^2}+\frac{{ch}^2}{{(h^2+k^2)}^2}\}=0\Rightarrow 2(h^2+k^2)+2(gh+fk)+c=0$
$\Rightarrow h^2+k^2+gh+fk+\frac{c}{2}=0$
Therefore required equation of locus is
$x^2+y^2+gh+fk+\frac{c}{2}=0$