Question

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

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

Answer 1

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

Let $P$ be the foot of $\perp$ from origin $O$ on any chord of the circle $S$ whose coordinates are $(\alpha ,\beta ).$

Then, the slope of $OP$ is $\frac{\beta }{\alpha }$ and thus the slope of chord is $-\frac{\alpha }{\beta }$ and its equation passing through $(\alpha ,\beta )$ is $y-\beta =-\frac{\alpha }{\beta }(x-\alpha )\Rightarrow \beta y-{\beta }^2=-\alpha x+{\alpha }^2$

$\Rightarrow \alpha x+\beta y={\alpha }^2+{\beta }^2$ ...(1)

Now, homogenizing the equation of the given circle

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

with the help of (1), we get

$x^2+y^2(2gx+2fy)\left(\frac{\alpha x+\beta y}{{\alpha }^2+{\beta }^2}\right)+c{\left(\frac{\alpha x+\beta y}{{\alpha }^2+{\beta }^2}\right)}^2=0.$ ...(2)

Now, equation (2) represents a pair of straight lines pass-ing through origin.

These lines will be at right angle if sum of the coefficients of $x^2$ and $y^2$ is zero.

i.e., ${({\alpha }^2+{\beta }^2)}^2+{({\alpha }^2+{\beta }^2)}^2+2g\alpha ({\alpha }^2+{\beta }^2)+2\beta f({\alpha }^2+{\beta }^2)+c({\alpha }^2+{\beta }^2)$

$\Rightarrow 2({\alpha }^2+{\beta }^2)+2g\alpha +3f\beta +c=0$ ...(3)

From equation (3), the locus of $P(\alpha ,\beta )$ is

$2(x^2+y^2)+2gx+2fy+c=0$

which is the required locus.