Solve $\frac{a+x-\sqrt{a^2-x^2}}{a+x+\sqrt{a^2-x^2}}=\frac{x}{b}$ for $x$ Can somebody please help me find $x$ in the following equation?

$$\dfrac{a+x-\sqrt{a^2-x^2}}{a+x+\sqrt{a^2-x^2}}=\dfrac{x}{b}$$
multiplication by $b$ and $a+x+\sqrt{a^2-x^2}$ and rearranging the variables we get
$$ab+bx-ax-x^2=\sqrt{a^2-x^2}(x+b)$$
now we obtain by squaring the equation $$2 x (a+x) \left(-2 a b+b^2+x^2\right)=0$$ which is easy to solve.
Hint:
$$ 
\frac{a + x - \sqrt{a^2 - x^2}}{a + x + \sqrt{a^2 - x^2}} = \frac xb
$$
$$
\frac{\sqrt{a+x} - \sqrt{a - x}}{\sqrt{a+x} + \sqrt{a - x}} = \frac xb
$$
$$
\frac{2x}{2a + 2\sqrt{a^2 - x^2}} = \frac xb
$$
$$
a + \sqrt{a^2 - x^2} = b
$$
$$
x = \pm \sqrt{a^2 - (b-a)^2} = \pm \sqrt{2ab - b^2}
$$