System of equations (Problem $50$ from $101$ algebra by Titu) 
Let $a$ and $b$ be given real numbers.
Solve the system of equations
$$\begin{aligned} \frac{x-y \sqrt{x^{2}-y^{2}}}{\sqrt{1-x^{2}+y^{2}}} &=a \\ \frac{y-x \sqrt{x^{2}-y^{2}}}{\sqrt{1-x^{2}+y^{2}}} &=b \end{aligned}$$
for real $x$ and $y$.

Solution -
Let $u=x+y$ and $v=x-y .$ Then
$$
0<x^{2}-y^{2}=u v<1, x=\frac{u+v}{2}, \text { and } y=\frac{u-v}{2}
$$
Adding the two equations and subtracting the two equations in the original system yields the new system
$$
\begin{aligned}
u-u \sqrt{u v} &=(a+b) \sqrt{1-u v} \\
v+v \sqrt{u v} &=(a-b) \sqrt{1-u v}
\end{aligned}
$$
Multiplying the above two equations yields
$$
u v(1-u v)=\left(a^{2}-b^{2}\right)(1-u v)
$$
hence $u v=a^{2}-b^{2} .$ It follows that
$$
u=\frac{(a+b) \sqrt{1-a^{2}+b^{2}}}{1-\sqrt{a^{2}-b^{2}}} \text { and } v=\frac{(a-b) \sqrt{1-a^{2}+b^{2}}}{1+\sqrt{a^{2}-b^{2}}}
$$
I did not get how they found values of $u$ and $v$ from
$u v=a^{2}-b^{2} .$ I mean obviously we can substitute the value in one of the equations and we will get some quadratic and we can find solution from there but the quadratic that I am getting is very large to handle, so is there some obvious step that directly lead solutions from $u v=a^{2}-b^{2}?$
Thank you

$$u-u\sqrt{uv} = (a+b)\sqrt{1-uv}$$
$$u = (a+b)\dfrac{\sqrt{1-uv}}{1-\sqrt{uv}}$$
Now substitute $uv=a^2-b^2$
