The tangent at a point $P$ on the curve $y=\ln(\frac{2+\sqrt{4-x^2}}{2-\sqrt{4-x^2}})-\sqrt{4-x^2}$ meets the $y-$axis at $T,$then find $PT^2.$
Let the point of tangency be $P(x_0,y_0)$ on the curve $y=\ln(\frac{2+\sqrt{4-x^2}}{2-\sqrt{4-x^2}})-\sqrt{4-x^2}$.
Equation of the tangent is $(y-y_0)=(\frac{dy}{dx})_{(x_0,y_0)}(x-x_0)$
I found $\frac{dy}{dx}=\frac{2\sqrt{4-x^2}+x^2}{x\sqrt{4-x^2}}$
Now the question has become difficult to simplify.If i put $x=0$ to find the coordinates of $T(0,\ln(\frac{2+\sqrt{4-x_0^2}}{2-\sqrt{4-x_0^2}})-\sqrt{4-x_0^2}-\frac{2\sqrt{4-x_0^2}+x_0^2}{\sqrt{4-x_0^2}})$ and the coordinates of $P$ are $(x_0,\ln(\frac{2+\sqrt{4-x_0^2}}{2-\sqrt{4-x_0^2}})-\sqrt{4-x_0^2})$.
If i use the distance formula now it is very difficult to simplify.Is there an efficient method to find $PT^2$?
This is not correct. $$\small\begin{align}\frac{dy}{dx}&=\frac{2-\sqrt{4-x^2}}{2+\sqrt{4-x^2}}\cdot\frac{(2+\sqrt{4-x^2})'(2-\sqrt{4-x^2})-(2+\sqrt{4-x^2})(2-\sqrt{4-x^2})'}{(2-\sqrt{4-x^2})^2}-\frac{-2x}{2\sqrt{4-x^2}}\\&=\frac{\frac{-4x}{\sqrt{4-x^2}}}{(2+\sqrt{4-x^2})(2-\sqrt{4-x^2})}+\frac{x}{\sqrt{4-x^2}}\\&=\frac{-(4-x^2)}{x\sqrt{4-x^2}}\\&=-\frac{\sqrt{4-x^2}}{x}\end{align}$$ Then, having $P(x_0,y_0), T\left(0,y_0+\sqrt{4-x_0^2}\right)$ gives $$PT^2=x_0^2+(4-x_0^2)=\color{red}{4}.$$