If $-1\leq x, y \leq 1$ and $x\sqrt{1-y^2} + y\sqrt{1-x^2} = 1$, find $x^2+y^2$

Question

Let $x,\, y\in\mathbb R,\ -1\leq x,\, y\leq 1$ such that $x\sqrt{1-y^2} + y\sqrt{1-x^2} = 1.$

Find the sum $S = x^2+y^2.$

Answer 1

Let $x=\sin\alpha$ and $y=\sin\beta$, where $\{\alpha,\beta\}\subset[-\frac{\pi}{2},\frac{\pi}{2}].$

Thus, the condition gives $\sin(\alpha+\beta)=1.$

Can you end it now?

Also, we can make squaring twice.

Indeed, the condition gives $$x^2(1-y^2)+y^2(1-x^2)+2xy\sqrt{(1-x^2)(1-y^2)}=1$$ or $$2xy\sqrt{(1-x^2)(1-y^2)}=1-x^2-y^2+2x^2y^2,$$ which after squaring again gives $$1+x^4+y^4-2x^2-2y^2+2x^2y^2=0$$ or $$(x^2+y^2)^2-2(x^2+y^2)+1=0$$ or $$x^2+y^2=1.$$

Answer 2

I think Michael's solution is much easier, but it's not something most people would think of. Instead, let's try doing this algebraically. First, I am going to isolate one of the terms on the left, so subtract by $y\sqrt{1-x^2}$: $$x\sqrt{1-y^2}=1-y\sqrt{1-x^2}$$ Now, when we square both sides, the square root on the left goes away: $$x^2(1-y^2)=1+y^2(1-x^2)-2y\sqrt{1-x^2}$$ Simplify some terms: $$x^2-x^2y^2=1+y^2-x^2y^2-2y\sqrt{1-x^2}$$ Add both sides by $x^2y^2$: $$x^2=1+y^2-2y\sqrt{1-x^2}$$ Subtract both sides by $1+y^2$ to isolate the square root on the right: $$x^2-y^2-1=-2y\sqrt{1-x^2}$$ Square both sides to get rid of the square root: $$x^4+y^4+1-2x^2y^2-2x^2+2y^2=4y^2(1-x^2)$$ Simplify some terms: $$x^4+y^4+1-2x^2y^2-2x^2+2y^2=4y^2-4x^2y^2$$ Add both sides by $4x^2y^2-4y^2$: $$x^4+y^4+1+2x^2y^2-2x^2-2y^2=0$$ Now, this is kind of hard to notice, but the $x^4+y^4+2x^2y^2$ terms kind of go together because that's $(x^2+y^2)^2$ and the $-2x^2-2y^2$ go together because that's $-2(x^2+y^2)$. If we make these substitutions, we get: $$(x^2+y^2)^2-2(x^2+y^2)+1=0$$ Substitute $S=x^2+y^2$: $$S^2-2S+1=0$$ Factor and solve: $$(S-1)^2=0\rightarrow S=1$$