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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.

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Hint: Look at each quadrant separately. –  mrf Jun 10 '12 at 11:56
    
sorry i did not get you –  meg_1997 Jun 10 '12 at 12:01
    
That is: first look at what happens when $x$ and $y$ are both positive; then, $x$ positive, $y$ negative; then, the othr way around; finally, both negative. –  Gerry Myerson Jun 10 '12 at 12:07
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Use $\LaTeX$ please, say, $\left\{(x,y)\big\vert(x-|x|)^2+(y-|y|)^2\le4\right\}$. –  Frank Science Jun 10 '12 at 12:12
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This is 36th question and I wonder why you don't pay attention to others who suggested to use $\LaTeX$. And as someone else pointed out in your previous question, please consider accepting an answer to your previous question if it's possible! –  Gigili Jun 10 '12 at 12:46
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1 Answer 1

up vote 2 down vote accepted

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 $y<0$:

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


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 $y \geq -\sqrt{1-x^2}$.

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That "final answer" looks very odd. The set is certainly not the whole thing framed in red. –  mrf Jun 10 '12 at 19:24
    
@mrf: Try Wolfram if you think it is "very odd". I checked it twice, found nothing wrong with it. Would be great to know what's "your final answer". –  Gigili Jun 10 '12 at 19:32
    
The picture is fine, but the inequalities you give are not. –  mrf Jun 10 '12 at 19:35
    
@mrf: I am not getting whats wrong with the above answer if it is will you please tell your answer –  meg_1997 Jun 10 '12 at 19:37
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The picture is correct as well as the text in the beginning of the last section, but the inequalities you put in a big red frame does not describe the set you show in the picture. Highlighting $x \ge -1$, $y \ge -1$ makes it look like those inequalities are the final answer. –  mrf Jun 10 '12 at 20:26
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