Formula of circle but with negative 2 in the exponent

[Boi]

I was messing around in Desmos when a strange thought came into my head: "What if i write formula for circle centred at point (h, k) but instead of 2 in (x-h)^2+(y-k)^2
I use -2?" So I wrote it and it got me a bit strange shape. Does anyone know how this shape is called? Is this also a section of some 3D solid? Please point me towards some paper about it if you know any
P.S. sorry for my bad english

 

[Otis]

[ (x – a)^-2 + (y – b)^-2 = r ^-2 ]

Does anyone know [what the graph's shape] is called?

Hi Boi. It can be viewed as a plot of two functions (y in terms of x). I don't know whether those curves have a special name, but the xy-graph is a plot of this form:

[imath]\quad y = \frac{\text{1st Quadratic Polynomial} \;±\; \sqrt{\text{Quartic Polynomial}}}{\text{2nd Quadratic Polynomial}}[/imath]

where the three polynomials' coefficients are in terms of a, b and r.

EG: The quartic polynomial is Ax⁴ + Bx³ + Cx² + Dx + E with
A = r²
B = -4ar²
C = 6a²r² – r⁴
D = -4a³r² + 2ar⁴
E = a⁴r² – a²r⁴

If not for the radical (in the form above), the two curves would be graphs of the sum and difference of two rational functions.

BTW: The asymptotic behavior continues forever, so the curves cannot form a cross-section of a 3D solid. We would need a closed curve, for that.
[imath]\;[/imath]

 

[Mr. Bland]

I would like to nominate the name "circsplosion" for this shape.

 

[Dr.Peterson]

I suggest "negellipse", by comparison with the hyperellipse and hypoellipse (also called superellipse).

 

[Boi]

Ok, but how do we turn it into negellipse? Now it's more like negcircle.

 

[Mr. Bland]

Ok, but how do we turn it into negellipse?

What you're tweaking at the moment is a circle in standard form:

[imath](x-h)^2+(y-k)^2=r^2[/imath]​

You've taken the exponent and changed it from [imath]2[/imath] to [imath]-2[/imath].

In order to ellipsify this, you'll need to start with the ellipse in standard form:

[imath]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1[/imath]​

Here, [imath]a[/imath] and [imath]b[/imath] are the radii of the ellipse. When [imath]a=b[/imath], you get a circle and the equation is equivalent to the first one up there.

 

[Dr.Peterson]

Ok, but how do we turn it into negellipse? Now it's more like negcircle.

A circle is a kind of ellipse.

Of course, I just made up the name, and don't care what you do with it.

 

[Boi]

So, I almost forgot about this random idea but then another weird idea knocked: what if i try to differentiate equation of hyperbola (half of the hyperbola actually)? I rushed to desmos, and put d over dx in front (half of) hyperbola eqution, and it got me this: seems familiar right? And with another half:

 

[Boi]

I probably wouldn't be that suprised by this connection if i knew how to differentiate square roots but right now i have no idea why there is any connection between changing sign of power in equation and differentiating half of that same equation