plotting the following set of points in the XY plane 2 :

Question

Represent the following set of points in the $XY$-plane

$$\left\{ (x,y) \big| (x-|x|)^2 + (y-|y|)^2 \leq 4 \right\}$$

Any help to solve this problem would be greatly appreciated. Thank you.

Answer 1

As I answered another question of your about absolutes, you have to consider possible cases:

• $x>0$ and $y>0$: In this case, we'll have $0 \leq 4$ which is always true so all the positive values of $x$ and $y$ are correct and a part of our final answer.
• $x>0$ and $y<0$: $$(2y)^2 \leq 4 \implies y^2 \leq 1 \implies -1 \leq y \leq 1$$

So the answer in this quadrant would be $x>0$ and $-1 \leq y \leq 0$

• $x<0$ and $y>0$: $$(2x)^2 \leq 4 \implies x^2 \leq 1 \implies -1 \leq x \leq 1$$ So the answer in this quadrant would be $y>0$ and $-1 \leq x \leq 0$

• $x<0$ and $y<0$:

$$4x^2 +4y^2 \leq 4 \implies x^2+y^2 \leq 1 \implies -1 \leq x \leq 1, -\sqrt{1-x^2} \leq y \leq \sqrt{1-x^2}$$

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To sum up the whole answers, from the first two cases, the union of the two sets is $x>0$ and $\{(y>0) \cup (-1 \leq y \leq 1)\}$ which is $y \geq -1$.

And from the other two cases, $\{(-1 \leq x \leq 0) \cup (-1 \leq x \leq 1)\} $ which is$-1 \leq x \leq 1$ and $\{(y>0) \cup (-\sqrt{1-x^2} \leq y \leq \sqrt{1-x^2})\}$ which is $y \geq -\sqrt{1-x^2}$.