# A simple equation with simple solution - how to solve it?

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NotEuler
I've been trying to solve y as a function of x in the following equation.
y-x=√(1-x^2)+√(1-y^2)

The equation look pretty simple, and according to Mathematica the solution is simple too: y=√(1-x^2)

The (proposed) solution looks so simple that I would guess there is some fairly straightforward way to get there. But I can't see it.
Is the only way squaring both sides until all roots are eradicated, and then solving whatever mess comes out of that? Or is there perhaps some nicer way?

## Answers and Replies

Whenever I see something like $\sqrt{1-x^2}$ I think of $x := \sin \theta$ or $x := \cos \theta$.

NotEuler
Thanks nuuskur. I had a look at your suggestion but still can't really see a way forward. Did you have a specific next step in mind?

NotEuler
I was wondering if symmetry could somehow be taken advantage of.
For example, if we look at the almost identical equation (first - changed to +)
y+x=√(1-x^2)+√(1-y^2)
which does not change in any way if the roles of x and y are reversed.
Does the symmetry of this equation help with solving it?

William Crawford
Whenever I see something like $\sqrt{1-x^2}$ I think of $x := \sin \theta$ or $x := \cos \theta$.
I agree. Therefore, let ##x = \sin\theta## and ##y = \sin\phi##, then the equation
$$y-x=\sqrt{1-x^2} + \sqrt{1-y^2}$$
becomes
\begin{align*} \sin\phi - \sin\theta &= \sqrt{1-\sin^2\theta} + \sqrt{1-\sin^2\phi} \\ &= \cos\theta + \cos\phi \end{align*}
or (by the sum-to-product trig identities)
$$\sin\Big(\frac{\phi-\theta}{2}\Big)\cos\Big(\frac{\phi+\theta}{2}\Big) = \cos\Big(\frac{\phi+\theta}{2}\Big)\cos\Big(\frac{\phi-\theta}{2}\Big)$$
and thus
$$\tan\Big(\frac{\phi-\theta}{2}\Big) = 1.$$
We can therefore conclude that
$$\phi - \theta = \frac{\pi}{2} + 2n\pi,\quad n\in\mathbb{Z}$$
and consequently that
\begin{align*} y &= \sin\phi \\ &= \sin\big(\theta + \frac{\pi}{2} + 2n\pi\Big) \\ &= \sin\big(\theta + \frac{\pi}{2}\Big) \\ &= \cos\theta \\ &= \pm\sqrt{1-\sin^2\theta} \\ &= \pm\sqrt{1-x^2}. \end{align*}

UPDATE
We have to be a little more carefull. Further considerations leads us to the conclusion that
$$y = \sqrt{1-x^2},$$
if and only if ##-1\leq x\leq 0##, is the only real solution to the original equation.

Last edited:
Delta2, Charles Link, mfb and 2 others
Mentor
An alternative without trigonometry:
##y-x=\sqrt{1-x^2}+\sqrt{1-y^2}##
##y-x-\sqrt{1-x^2}=\sqrt{1-y^2}##
square
##y^2+x^2+(1-x^2) - 2xy - 2y\sqrt{1-x^2} +2x\sqrt{1-x^2} = 1-y^2##
collect equal terms:
##2y^2 - 2xy - 2y\sqrt{1-x^2} +2x\sqrt{1-x^2} = 0##
simplify more:
##y^2 + y(-x-\sqrt{1-x^2}) +x\sqrt{1-x^2} = 0##
You can solve it like a regular quadratic equation from here on, but remember that the absolute term is the product of the two roots while the linear term is the negative sum of them. And that's exactly the pattern we have here so we can just read off the two intermediate solutions:
##y=x##
##y=\sqrt{1-x^2}##
Plugging these back into the original equation tells us that y=x is not a solution (it was added by squaring the equation), but the other one is real.

archaic and Charles Link
NotEuler
Thanks for your input everyone. Very happy to see there are at least two reasonably 'clean' ways to solve it.

I originally tried something similar to mfb's solution, but I think my mistake was having both square roots on one side and the other terms on the other side. Seems it's much better to move √(1-y^2) to one side and everything else to the other, and proceed from there.

Homework Helper
Gold Member
I got a solution by grouping the x terms on one side, and the y terms on the other, and squaring both sides twice. The result, after one algebraic step is ##y^2-x^2=y^4-x^4=(y^2-x^2)(y^2+x^2) ##. One solution that emerges is ## x^2+y^2=1 ##, which is the same solution found by the others above. Further analysis, as in the Update in post 5, shows the unit circle in the second quadrant is indeed the solution, with the other quadrants failing to satisfy the original equation.

Delta2 and hutchphd
Fred Wright
Taking mfb's suggestion:$$y-\sqrt{1-y^2}=x+\sqrt{1-x^2}$$Squaring both sides and cancelling terms,$$-y\sqrt{1-y^2}=x\sqrt{1-x^2}$$Squaring both sides leads to,$$y^4-y^2+x(1-x^2)=0$$$$y^2=\frac{1}{2}\pm \sqrt{x^4-x^2+\frac{1}{4}}$$observe$$x^4-x^2+\frac{1}{4}=(x^2-\frac{1}{2})^2$$ and so,$$y^2=\frac{1}{2}\pm( x^2-\frac{1}{2})$$$$y=\pm \sqrt{\frac{1}{2}\pm( x^2-\frac{1}{2})}$$Plugging these four roots into the original equation we find that the equation is only satisfied at the four points ##(1,1)##, ##(-1,-1)##, ##(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})##, ##(-1,0)##.

Delta2
Homework Helper
Gold Member
Perhaps I am missing something, but from what I could determine, the solution ## x^2+y^2=1 ## will hold anywhere in the second quadrant. It will not work for the points (1,1) and (-1,-1) which come from ## y=x ## and are extraneous.
Edit: I must be missing something, because I see that these two points do indeed work. Perhaps @Fred Wright has found something.
Suggestion: Try doing a graphical solution= graph ## z =y-\sqrt{1-y^2} ## vs. ##y ##, and ##z=x+\sqrt{1-x^2} ## vs. ##x ##, with common ##x ## and ## y ## on the abscissa, and see where points exist where the z's are equal. It appears our solutions may be correct, and that ## (1,1) ## and ## (-1,-1) ## also work as graphical oddities. (if you sketch the graphs, you should see why I called it a graphical oddity).

Last edited:
Delta2 and mfb
Fred Wright
Perhaps I am missing something, but from what I could determine, the solution ## x^2+y^2=1 ## will hold anywhere in the second quadrant.
Your absolutely correct, Charles. I'm embarrassed and have no excuse for my stupid four points nonsense except for a possible lapse into dementia.

Homework Helper
Gold Member
Still a plus for you @Fred Wright =You spotted two points that the rest of us missed. It sent me back to the drawing board, and the graphs show how those two points ##(1,1) ## and ## (-1,-1) ## get in there.

Homework Helper
Gold Member
@QuantumQuest @fresh_42 This is kind of a rather interesting math problem. It might be a good high school problem for a challenge 6 months from now. :)