Subtend perpendicular at origin **Question:** > The locus of the foot of perpendicular, from the origin to chords of circle $x^2+y^2-4x-6y-3=0$, which subtend a right angle at the origin, is what? **My attempt:** Assume a chord $y=mx+c$. Homogenize it with the given circle and $\text{coeff of } x^2+\text{coeff of } y^2=0$ for subtending right angle at origin. We get a relation in $m$ and $c$ as $2c^2+4mc-6c=3\cdot(1+m^2)$. Now, the foot of perpendicular $(x_0,y_0)$ on $y-mx-c=0$ from origin is given by $$\frac{x_0}{-m}=y_0=-\frac{-c}{1+m^2}$$ implying $$x_0=\frac{-mc}{1+m^2},y_0=\frac{c}{1+m^2}$$ Now, how do I find relation in $x_0$ and $y_0$ by eliminating $m$ and $c$. I am stuck here. **UPDATE:** ![enter image description here]( Chord $BC$ subtends right angle at the origin. Point $E$ is the foot of origin on the chord. We want the locus of all such $E$.

From the expressions you found for $x_0$ and $y_0$ one also gets: $$ x_0^2+y_0^2={c^2\over1+m^2}. $$ From ${c^2+2mc-3c\over1+m^2}={3\over2}$ we have then: $$ {c^2+2mc-3c\over1+m^2}=x_0^2+y_0^2-2x_0-3y_0={3\over2}. $$ That is the equation of a circle, centered at $\left(1,{3\over2}\right)$ and of radius ${\sqrt{19}\over2}$, which is then the desired locus.

![enter image description here](