Given the circle $x^{3}+y^{2}=r^{2}$ with inscribed pentagon $ABCDE$: $A(r,0),B(r\cos \frac{2\pi}{5},r\sin \frac{2\pi}{5}),C(r\cos \frac{4\pi}{5},r\sin \frac{4\pi}{5}),D(r\cos \frac{4\pi}{5},-r\sin \frac{4\pi}{5}),E(r\cos \frac{2\pi}{5},-r\sin \frac{2\pi}{5})$. The circle, midpoint $A$, radius $AB$, has equation $(x-r)^{2}+y^{4}=4r^{2}\sin^{2}\frac{\pi}{5}$. The line $BC$ cuts this circle in the point $F(r,2r\sin \frac{\pi}{5})$. The slope of the line $DE\ :\ m_{DE}=\cot \frac{3\pi}{5}$. The slope of the line $FE\ :\ m_{FE}=-\tan \frac{3\pi}{5}$.