I have to show that if we draw the derivatives $d\phi/ds$ from one point $P$ of the scalar field in every direction, the ends of the line segments representing the value of $d\phi/ds$ will lie on a circle (or a sphere, but I'm limiting the problem to 2D). $\phi$ is an equiscalar surface.
I've used law of cosines to define $a$ (in the figure): $$a^2=(\nabla\phi/2)^2+(\frac{d\phi}{ds})^2-\nabla\phi\frac{d\phi}{ds}\cos{\alpha}$$ I would now like to show that $a=const$ but I don't know how. I've tried choosing different direction $\hat{e}$ and obtaining another equation from law of cosines but it seemed pointless.
I've finally figured this out and thought of sharing. The main trick is to write $d\phi/ds$ as $$d\phi/ds=\nabla\phi\cdot\hat{e}=\nabla\phi\cos{\angle\nabla\phi,\hat{e}}$$ So we have $$a^2=(\nabla\phi/2)^2+(d\phi/ds)^2−\nabla\phi d\phi/ds\cos{\alpha}$$ $$a^2=(\nabla\phi/2)^2+\nabla\phi^2\cos^2{\alpha}−\nabla\phi \nabla\phi\cos{\alpha}\cos{\alpha}$$ $$a^2=(\nabla\phi/2)^2+\nabla\phi^2\cos^2{\alpha}−\nabla\phi^2\cos^2{\alpha}$$ $$a^2=(\nabla\phi/2)^2$$ $$a>0$$ $$a=(\nabla\phi/2)=const$$