Suppose we have a smooth differentiable planar curve given as $F(x,y)=0$. Suppose that the centroid of the curve $F=0$ is $F(x_0,y_0)$.
My question is:- $$\nabla F(x_0,y_0)=0$$ That is, does the gradient of $F$ becomes zero at $(x_0,y_0)$?
Now some examples supporting this statement are circles and ellipses.
I would be grateful if anyone proves the statement or gives a counter example. Also I welcome anyone who would like to generalise this statement in any direction.
Thanks!
Let $$F(x,y) = x^4 - 4 x^3 + 2 x^2 y^2 - 4 x y^2 + 16 x + y^4 - 16.$$
The graph of $F(x,y) = 0$ is a circle of radius $2$ centered at the origin, so its centroid is $(x_0,y_0) = (0,0).$
But
$$ \nabla F(x,y) = (4 (x^3 - 3 x^2 + x y^2 - y^2 + 4), 4 y (x^2 - 2 x + y^2)),$$
which is not zero at $(x,y)=(0,0).$
(To obtain this function, I treated the coordinates $x,y$ as the real and imaginary parts of a complex number $z = x+iy,$ applied the Möbius transformation $z \mapsto (z+1)/(z-1),$ separated the result into real and imaginary parts $x'$ and $y'$, and applied the circle formula $x'^2 + y'^2 = 4.$ This produced a circle with center at $(-1,0)$, so I changed $x \to x-1$ to translate the circle's center to the origin, then expanded the polynomial. I was hoping that the Möbius transformation, while it mapped a circle to a circle, would distort things enough to put the zero gradient off-center, and this turned out to be a good guess.)