Conflict in the graph of $\{(x,y)|(x-|x|)^2+(y-|y|)^2 \leq 4\}$ directly and by cases

Question

Question

$\text{Plot the graph of}$ $\{$ $(x,y) \in \mathbb{R}$ | $(x-|x|)^2+(y-|y|)^2 \leq 4\}$.

My attempt

• Case $1$ : $x,y<0$ $⇒(x+x)^2+(y+y)^2\leq4⇒4x^2+4y^2\leq4⇒x^2+y^2\leq1⇒ $ A shaded circle with center at origin and radius $=1.$

• Case 2 : $x>0,y<0⇒ 4y^2\leq4⇒y^2\leq1⇒|y|\le1⇒y=1,-1⇒$ Shaded region from $-1$ to $1$ on $Y-$axis.

• Case 3: $x<0,y>0⇒ $ Similar to above shaded from $[-1,1]$ on $X-$axis.

• Case 4 : $x,y>0⇒(x-x)^2+(y-y)^2\le4⇒$ All $+\text{ve}$ Real Numbers.

So when I plot this all the region should be the answer.

Correct Answer

But when I plot above inequality directly ,

Problem

Where is it wrong? Because the final graph will be union of all the cases right?

References

Link for interactive graph of my cases (image 1): https://www.desmos.com/calculator/cwhuffantk

Link for the graph of the original inequality (image 2): https://www.desmos.com/calculator/bwc19xvibn

Answer 1

Case $1$ : $x,y<0$ $⇒(x+x)^2+(y+y)^2\leq4⇒4x^2+4y^2\leq4⇒x^2+y^2\leq1⇒ $ A shaded circle with center at origin and radius $=1.$

Correct, and note that Case 1 lies only in Quadrant 3. (So, restrict the shaded red circle to just the portion in Quadrant 3.)

Case 2 : $x>0,y<0⇒ 4y^2\leq4⇒y^2\leq1⇒|y|\le1⇒y=1,-1⇒$ Shaded region from $-1$ to $1$ on $Y-$axis.

Yes, and Case 2 lies in Quadrant 4.

Case 3: $x<0,y>0⇒ $ Similar to above shaded from $[-1,1]$ on $X-$axis.

Yes, and Case 3 lies in Quadrant 2.

Case 4 : $x,y>0⇒(x-x)^2+(y-y)^2\le4⇒$ All $+\text{ve}$ Real Numbers.

Note that $$0\le4\kern.6em\not\kern-.6em\implies x>0$$ (for example, notice that $x=-3$ satisfies the LHS, but not the RHS, of the implication); rather, $$0\le4\implies (x,y)\in\mathbb R^2.$$ But since case 4 lies in Quadrant 1, we just want the the entirety of Quadrant 1.

Finally, taking the union of the four cases (and considering the points on the two axes) gives the required region https://www.desmos.com/calculator/spznk8ziya, which matches what Desmos directly gave.

Answer 2

Why is there a difference in the two solutions?

The issue here is that you haven't considered the restrictions on x and y in each scenario. In each case you assume that x and y are either positive negative or a combination of the two. However, in your graph, you assume that each case holds simultaneously for all values x and y (when they only hold in the case where x and y meet the required restriction).

How to resolve this difference?

Let's look at the four claims you make.

Case 1: you are considering a shaded circle of radius 1. However, we assume that $x<0$ and $y<0$, so clearly we don't have a full circle here, we only have one quarter of a circle (where x and y are both negative). We therefore, must only sketch the circle for the bottom left quadrant.

Case 2: by the same reasoning, we are only consider $|y|\le 1$ in the bottom right quadrant of the graph as this only holds under the assumption that x is positive and y is negative.

Case 3: again, as above, $|x| \le 1$ only applies in the top left quadrant by the same reasoning. We see that this only holds when we assume that x is negative and y is positive.

Case 4: as before, all pairs of real numbers $x,y$ only applies in the top right quadrant (positive x, and positive y). Since we make this assumption to come to that conclusion, it isn't correct to consider this to hold for values outside positive x and positive y combinations.

The main take away here, is that these 4 cases only give you those equations if you make an assumption about the values of x and y. Therefore, if you apply these graphs to the entire cartesian plane, this does not follow on from the cases. Each one applies in a different quadrant and so you need to look at the suitable values of x and y in each case and shade each quadrant appropriately.

If you do this, then you will arrive at the same answer as the one in the Desmos produced graphic that you have included in your question.