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?
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.