已知 \(x\sqrt{1-y^2}+y\sqrt{1-x^2}=1\)，求证 \(x^2+y^2=1\)。
\[x\sqrt{1-y^2}+y\sqrt{1-x^2}=1
\]
\[x\sqrt{1-y^2}=1-y\sqrt{1-x^2}
\]
\[x^2(1-y^2)=1+y^2(1-x^2)-2y\sqrt{1-x^2}
\]
\[2y\sqrt{1-x^2}=1+y^2-x^2
\]
\[4y^2-4y^2x^2=1+y^4+x^4+2y^2-2x^2-2x^2y^2
\]
\[x^4+y^4-2y^2-2x^2+2x^2y^2+1=0
\]
\[{(x^2+y^2)}^2-2(x^2+y^2)+1=0
\]
\[{(x^2+y^2-1)}^2=0
\]
\[x^2+y^2=1
\]