Sketching level curves for $f(x,y)=x/(x^2+y^2)$

Question

I'm trying to sketch level curves for $f(x,y)=x/(x^2+y^2)$ at $c=-2,0,4$. If $c=0$ we know that $C_0=\{(x,y)\in \mathbb{R}^2 \setminus (0,0):x=0\}$. The other cases are a little tougher.

For $c=-2$ we can solve

$$\frac{x}{x^2+y^2}=-2 \iff y= \pm \sqrt{-x^2-x/2}, x\in [-1/2,0]$$

But I don't know how to sketch that. Alternatively, in polar

$$\cos \theta= -2r$$

But how do I know which are the bounds on $\theta,r$? Also, how do I sketch that curve? I'm really lost.

Answer 1

hint

for $c=-2$.

$$\frac{x}{x^2+y^2}=-2 \iff$$

$$x^2+y^2+\frac{x}{2}=0 \iff$$

$$(x+\frac 14)^2+y^2=\frac{1}{16}$$ and this represents a circle whose center is $(\frac{-1}{4},0)$ and radius $\sqrt{\frac{1}{16}}=\frac 14$.

Answer 2

$$f(x,y)=c \Longleftrightarrow \frac{x}{x^2+y^2}=c$$ Since $(x,y)\neq(0,0)$ $$x=cx^2+cy^2$$ $$0=cx^2-x+cy^2$$ If $c\neq 0$ $$0=x^2-\frac{x}{c} +y^2$$ Completing the square $$\frac{1}{4c^2}=\left(x-\frac{1}{2c}\right)^2 +y^2$$ The level curve is a circle centered in $\left(\frac{1}{2c},0\right)$ and of radius $\left\lvert\frac{1}{2c}\right\rvert$