Class of lines $x^n+y^n=r^n$ Playing around with circles led me to the following observations:
For the equation $x^n+y^n=r^n$ and any real $r$:


*

*For any positive real $n$, there seem to be three possible lines that can be drawn: rounded squares, the top-right (bottom-right for a negative $r$) corner of a rounded square, and a negatively sloped line with a bulge in it around the origin that resembles the top-right (bottom-right for negative $r$) corner of a rounded circle. $r$ can change without affecting the shape of the curve; all it affects is the x- and y-intercepts (always equivalent to $r$).

*When $n$ is a negative real, the options are inverted: imagine drawing a square and extending the lines outward to ±∞ on both axes; these are the asymptotes for the curves drawn, and while sometimes there are four curves, other times there are two, and other times just one. 


Pictures:

$n=3$

$n=4$

$n=4.3$

$n=-3$

$n=-4$

$n=-4.3$
For integers the pattern is pretty straightforward: the first type for an odd $n$, while the second for an even one. The third type never appears for an integer $n$. 
What I can’t figure out is the pattern when $n$ is not an integer; there doesn’t seem to be a pattern. The odd/even thing fails. 
Is this a known phenomenon? Is there a way to predict what the line will look like given...something about $n$?
(As an aside, posting a polar formula equivalent to these would be very helpful, possibly.)
 A: Using spherical coordinates, you have $\cos^n\phi + \sin^n\phi = 1$. For negative $n$, write $m = -n$ to give $\cos^m\phi + \sin^m\phi = \cos^m\phi \cdot \sin^m\phi$ which explains the obvious difference between positive and negative $n$.
The reason why for non-integer values there are problems in other than the first quadrant can be explained by trigonometric features of the angular functions. 
For example, take the negative angle.  Since $\cos (- \phi) = \cos \phi $  and $\sin (- \phi) = - \sin \phi $ you get 
$\cos^n(-\phi) + \sin^n(-\phi) = \cos^n\phi + (-1)^n \sin^n\phi = 1$. For $n = \pm 4$, as in your examples, this means that the solution in the first quadrant will be mirrored in the third. However, for non-integer $n$, you are facing $(-1)^n = \exp (i \pi n)$ which in general is complex, so there will be no real solution at all.  The same arguments hold for an angular  shift of $\pm \pi/2$.
This explains why you have only a solution in the first quadrant for non-integer $n$. 
Other than that, the shape of the curve smoothly varies in the first quadrant as you change $n$. For example, your curves for $n=4.3$ will lie in between those for   $n=4$ and $n=5$ in the  first quadrant.
