How to define a curve surrounding the origin?

Question

Is there any criteria for stating that a curve is sorrounding the origin?

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For example:
Consider $\Bbb D=\{x^2+y^2-y\le\sqrt{x^2+y^2}\}$, that is in polar coordinates: $\Bbb D_{(\rho, \theta)}=\{0\le\theta\le2\pi,\, 0\le\rho\le1+\sin{\theta}\}$. In particular it's $\{0\le\theta\le2\pi\}$ because the curve $\{x^2+y^2=y+\sqrt{x^2+y^2}\}$ surrounds the origin; and I stated that thinking "it is the locus of points $(x,y)$ which are away from the origin $\left(y+\sqrt{x^2+y^2}\right)^{\frac{1}{2}}$".

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Now, I think that a similiar condition could be a right answer in these cases; but is there a more general condition about?

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[EDIT]
Maybe the true question was: "when a curve given into cartesian form can be written into $\{0 \le \theta \le 2\pi, 0 \le \rho \le f(\theta)\}$ for some function $f$"?

Answer 1

This will probably not be the answer you are looking for. I don't think a correct answer of the kind you are looking for exists. But explaining this requires more than will fit into comments.

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Consider this simple closed curve that contains the origin.

It is the algebraic curve given by $\left((x-1)^2+(y-.3)^{2}+1\right)^2-4(x-1)^2=1.03$, which is an oval of Cassini whose center has been displaced from the origin.

Notice that the curve crosses the positive $x$-axis three times.

This curve and its interior include the origin, so it is true that in every direction $\theta \in [0,2\pi]$ from the origin there is some point of the curve. There are even three points in some directions.

It is also true that the curve encloses a finite region of the plane. But you will not be able to describe that region as $\mathbb D_{(\rho, \theta)}=\{0\le\theta\le2\pi,\, 0\le\rho\leq f(\theta)\}$ for any function $f.$

If what you want is a region that can be described in the form $\{0\le\theta\le2\pi,\, 0\le\rho\leq f(\theta)\}$ for some function $f,$ why not just say so? Why do you need to state any other conditions?

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If you don't actually care about writing a domain in such a form (that is, if the domain in the question was not actually relevant to the question), and all you actually want is that a curve be simple, closed, and contain the origin either in the interior of the curve or on the curve itself, why not just say so? Why do you need to state any other conditions?

Merely knowing that a curve is the locus of points $(x,y)$ that are at a distance $(p(x,y))^{1/2}$ from the origin for some polynomial $p$ is not enough to conclude that the curve contains the origin. For example, the boundary of the region shown below is a curve that is the locus of points $(x,y)$ that are $\left(x^2+y^2-2y+\frac12\right)^{1/2}$ away from the origin, but it clearly does not contain the origin.

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In summary, don't make things pointlessly complicated.

Answer 2

For a circle constant $C$ also known as power should be negative ( zero if on circle)

$$ x^2+y^2 + a x + b y + C = 0$$

So also an ellipse will enclose the origin when $C<0$ in:

$$a x^2 +2h x y+b y^2+2 f x+2 g y + C=0. $$

Considering pole/polar of an oval, "tangent length" from origin is imaginary.

This is easier with differential calculus than with analytical geometry, if allowed here.

An oval curve polar coordinates $ r= f(\theta) $ has derivative $ r^{'}$. If

$$\sin \psi= \dfrac{r}{\sqrt{r^2+r^{'2}}} $$

is positive everywhere then the curve contains/encloses the origin. The $\psi$ is radius vector to polar curve angle.

Polar plots of a Newton ellipse ( semi latus-rectum $p=1$, eccentricity $e= 0.6$ )

$$\dfrac{p}{r}= 1- e \cos\theta $$

Differentiate with respect to arc $s\; \frac{dr}{ds}=\cos \psi , \frac{d\theta}{ds}=\frac{\sin \psi}{r}$ we get

$$\tan\psi= \frac{\cos \theta-1/e}{\sin \theta}$$

$\sin \psi$ is everywhere positive and the polar plot also seen to contain the focus or origin

Basically we should use differential equations to find how the pole aspects portions of a closed loop curve. Whether pole is included or excluded in a loop can be seen by intersection of $(\sin \psi, s)$ curves with arc-length x-axis.We are concerned with the roots.

Behaviour of first curve is like Newton's ellipse where $\sin \psi$ is always positive.

The second curve behaves differently with alternating arc segments with positive and negative $\sin \psi$ which can be exemplified using the eccentric circle ( chosen because it is the simplest closed curve) given below with two tangent points.

Property of Circle product of segments $ OA \cdot OB=T^2 $ can be written:

$$ 2a \sin \psi= (1-T^2/a^2) $$

The appearance of points of tangency demarcating regions of sign change of $\sin\psi$ is in itself a proof of the pole existing outside the loop of curve. The situation of pole/polar is evident for external pole.

EDIT1:

Two example where curvature involves $\sin \psi$ are given taking the above further :

Eccentric Circles : $n=1$; Constant curvature $$ \dfrac{d\psi}{ds} +\dfrac{(\sin \psi)^1}{r} = \pm 1/a; \; $$ Precessing Orbits:$\; n=2 $ $$ \dfrac{d\psi}{ds} +\dfrac{(\sin \psi)^n}{r} = \pm 1/a; \; $$

The plots show how in two precession orbits cases $\sin \psi $ influences inclusion or exclusion of the pole/origin.

Answer 3

Using complex number as coordinates:

\begin{align} z &= re^{i\theta} \\ dz &= (r'+ri) e^{i\theta} d\theta \end{align}

By Cauchy integral formula:

$$f(a)=\frac{1}{2\pi i}\oint_C \frac{f(z)}{z-a} \, dz$$

where $C$ is a simple closed curve enclosing $a\in \mathbb{C}$ in anti-clockwise sense.

In particular, take $a=0$ and $f(z)=1$ with contour $C=\partial D$ (boundary of an open set $D$),

$$\oint_{\partial D} \frac{dz}{z}= \begin{cases} 2\pi i, & 0\in D \\ 0, & 0\notin D \end{cases}$$

You may regard this as two-dimensional Gauss' Law. Please feel free to refer my two-dimensional treatment on electric field in Physics Stack Exchange.

See also winding number, which is relevant to simple closed curve, for your further interest.