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

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:
$$(S-1)^2=0\rightarrow S=1$$
