Locus of foot of perpendicular

Question

A circle of radius $r$ passes through the origin $O$ and cuts the axes at $A(a,0)$ and $B(0,b)$. What is the locus of the foot of perpendicular from $O$ to $AB$?

I found the equation of circle passing through $A$, $B$ and $O$ and then found $k/a=h/b$ (taking the foot as $(h,k)$). I also found that $a^2+b^2=4r^2$. What's next?

Why do you refer to a locus? What is variable here? The centre of the circle through OAB is the midpoint of AB, btw. — David Quinn, Sep 13 '15 at 13:41
a,b are the variables here.r is the only constant @DavidQuinn — , Sep 13 '15 at 13:43
I don't understand why you introduced a circle. The foot of the perpendicular from $O$ to $AB$ only depends on the point $O$ and the line $A $? — mathcounterexamples.net, Sep 13 '15 at 13:48
@mathcounterexamples.net that's what the question in front of me says.Thats all ! — , Sep 13 '15 at 13:49
Source http://www.flipkart.com/mathematics-jee-advanced-english-1st/p/itmdmtb7yrngbfrg @mathcounterexamples.net — , Sep 13 '15 at 13:51

score 1 · Answer 1 · answered Sep 13 '15 at 13:58

Denote by $\theta$ the polar angle of the perpendicular. Then $a = 2r\sin \theta$, $b= 2r\cos\theta$, and the length of the perpendicular is $r |\sin 2\theta|$. So the locus equation in polar coordinates is $\rho = r|\sin 2\theta|$, which gives a nice flower.

Locus of foot of perpendicular

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