The circle has the uniform shape because a second derivative is 1. That is an intuitive guess - the line turns around at constant rate (i.e. the first derivative changes at constant rate), which means that it is not dependent on x and y coordinates. If the rate of the turn would increase, one would get inward spiral, etc. The shape of the circle is uniform so the step of the turn in x variable and in y variable is the same.
How to support this mathematically? Should a divergence be computed? I tried simply computing derivates however the circle is not a function, it is an equation and it is quickly very confusing.
One can parametrize the circle by$$\gamma:[0,2\pi]\to\mathbb{R}^2,\quad t\mapsto(\cos t,\sin t).$$Since for every $t$ we have$$\|\dot{\gamma}(t)\|=\|(-\sin t,\cos t)\|=1,$$this is an arc length parametrization. Hence, the curvature at a point is given by the norm of the second derivative. But$$\|\ddot{\gamma}(t)\|=\|(-\cos t,-\sin t)\|=1$$is constant, and so the circle has a constant curvature.