Why does Desmos not plot $ \sqrt{\frac{(x+y)^2}{2}} = \sqrt{x^2+y^2} + \sqrt{2} $?

Question

I am working on the following problem:

I need to find the points above the line $l:y=-x$ for which it holds that the distance to $l$ is $\sqrt{2}$ greater than the distance to the origin.

I worked out a way to calculate the distance from a point to the line $l$: $$ \sqrt{\frac{(x+y)^2}{2}}. $$ And then obviously the distance from a point to the origin is: $$ \sqrt{x^2+y^2}. $$ I then tried to plot the function: $$ \sqrt{\frac{(x+y)^2}{2}} = \sqrt{x^2+y^2} + \sqrt{2} $$ in Desmos and wolframalpha and both plots gave me nothing...
I even tried to plot just: $$ \sqrt{\frac{(x+y)^2}{2}} = \sqrt{x^2+y^2} $$ for which I know the line $y=x$ should be a solution, but this didn't plot anything either.

I'm not sure what is causing this. Why are Desmos and wolframalpha not able to Plot the curves ? Any ideas?
Many thanks,
Hugo

Answer 1

PROBLEM :

There is 1 MAJOR Issue ( & 1 MINOR Issue ) with your formulation , which is why neither DESMOS nor Wolfram will give the necessary graph.

(MAJOR) Distance to Origin ( $A$ ) is always more than ( or Equal to ) Distance to the line ( $B$ ) , in the given Case.
Hence , when you add $\sqrt{2}$ to $A$ , you are making it even larger than $B$ & there is no Point which will work out.

You should rather add $\sqrt{2}$ to $B$ to make the Correct formulation.

This Image highlights that Issue :

(MINOR) You are using $\sqrt{(x+y)^2}$ when trying to get $B$ magnitude , which is a little round-about.
Instead , you should use the More Direct $|x+y|$ & check that.

OUTPUT :

When I made those Changes , I got the necessary graph :

I used Wolfram Online Tool here. DESMOS too will give that Plot without issues now.

Answer 2

The equation you want to plot seems to be $\sqrt{x^2+y^2}=\frac{|x+y|+2}{\sqrt2}$ which should scream parabolas with focus $(0,0)$ and directrices $x+y+2=0$ and $-(x+y)+2=0$ at you.

Repeated squaring indeed yields $\frac14(x^2+y^2-\frac{(x+y+2)^2}2)(x^2+y^2-\frac{(x+y-2)^2}2)=0$ which plots as two parabolas.

However, squaring can add solutions, and the parts around the vertices $(\pm\frac12,\pm\frac12)$ bounded by the intersection with the line $|x+y|=0$ are not on the original plot, precisely because those belong to $\sqrt{x^2+y^2}=\frac{|x+y|-2}{\sqrt2}$ (from the title) which doesn't have points as John Wayland Bales shows in a comment to the question.