I can't solve this basic equation?

Question

I need to make either $y$ or $x$ the subject of this equation but I can't seem to get either one

$\frac{2x}{1+x^2+y^2}+ \frac{1}{\sqrt2} =0$

I always end up with

$-2\sqrt2x-1-x^2=y^2$ which is obviously not going anywhere...

Answer 1

Note that

$$x^2+2\sqrt2x+1+y^2=0\iff (x+\sqrt 2)^2+y^2=1.$$ This is the equation of a circle with center $(-\sqrt 2,0)$ and radius $1.$

We have $$y=\pm\sqrt{1-(x+\sqrt 2)^2}$$ and $$x=-\sqrt2\pm\sqrt{1-y^2}.$$

Answer 2

What is wrong with seeing it as a quadratic in $x$ or $y$ and feeding it to the quadratic formula? The simplest is $$y=\pm\sqrt {-x^2-2\sqrt 2 x -1}\\ y=\pm\sqrt{1-(x+\sqrt 2)^2}$$ which shows $x$ must be in the range $[\sqrt 2-1,\sqrt 2+1]$

Answer 3

$-2\sqrt2x-1-x^2=y^2$ which is obviously not going anywhere...

Why do you say that?

It looks to me like that is "obviously" going to

$y = \pm \sqrt{-2\sqrt 2 x - 1 - x^2}$

which can be solved by the quadratic equation to:

$y = \pm \sqrt{ - (x - \frac {2\sqrt 2 - \sqrt{8 - 4}}{-2})(x - \frac {2\sqrt 2 + \sqrt{8 - 4}}{-2})}$

$y = \pm \sqrt{-(x + \sqrt 2-1)(x+\sqrt 2 + 1)}$

As $-(x + \sqrt 2-1)(x+\sqrt 2 + 1)$ must be greater than or equal to zero we must have $-1 -\sqrt 2 \le x \le 1- \sqrt 2$.

....

But perhaps simpler would be noting that $y^2 + x^2 + 2\sqrt {2} x + 1= 0$ is

$y^2 + x^2 + 2\sqrt{2}x + 2 = 1$ so

$y^2 + (x+\sqrt 2)^2 = 1$

So this is the graph of a circle and expressible as either $y = \pm \sqrt{1- (x+\sqrt 2)^2}$ or as $x = -\sqrt 2 \pm \sqrt{1 -y^2}$