Compute the real value $k$ for which the graph of $x^k+y^k=1$ is the same as the lower half of $(x-1)^2+(y-1)^2=1$ over $(0,1)$ or show that no such $k$ exists.
I came up with this while daydreaming. My approach was to start out on Desmos testing values of $k$- I got $k\approx 0.5464$, but there were still some slight divergences. Then I decided to test whether such a $k$ existed by looking at the point on $y=x$ that would have to be shared by both. The point on the circle is $(1-\frac{\sqrt2}{2},1-\frac{\sqrt2}{2})$ and the point on the other curve is where $2x^k=1$, or $x=y=\big(\frac{1}{2}\big)^k$. Then we need that $\big(\frac{1}{2}\big)^k=1-\frac{\sqrt{2}}{2}$, or (after a bit of simplifying) $k=\log_{1-\frac{\sqrt{2}}{2}} \big(\frac{1}{2}\big)\approx 0.5645$. I went on Desmos and plugged this in but found that there still seemed to be very slight differences between the two. I would think (assuming Desmos is right) that this implies that no such $k$ exists. Is there any flaw in my logic?
Can $K$ be $2$ to make it work? Because $\cos(a)\cdot\cos(a) + \sin(a)\cdot\sin(a) = 1$, and you can replace $x$ with $r\cdot\cos(a)$, and replace $y$ with $r\cdot\sin(a)$. and $r = 1$ according to the circle function.