Minimum value of $\frac{b+1}{a+b-2}$ 
If $a^2 + b^2= 1 $ and $u$ is the minimum value of the $\dfrac{b+1}{a+b-2}$, then find the value of $u^2$. 

Attempt: 
Then I tried this way: Let $a= bk$ for some real $k$. 
Then I got $f(b)$  in terms of b and k which is minmum when $b = \dfrac{2-k}{2(k+1)}$ ... then again I got an equation in $k$ which didn't simplify. 
Please suggest an efficient  way to solve it.

Try with $b=\cos 2x$ and $a= \sin 2x$.
\begin{eqnarray}{b+1\over a+b-2}&=& {2\cos^2 x\over  -\cos^2x+2\sin x \cos x -3\sin^2x}\\
&=& {2\over  -1+2\tan x  -3\tan^2x}\\
&=& {2\over  -1+2t  -3t^2}
\end{eqnarray}
where $t= \tan x $. So the expression will take a minimum when quadratic function $g(t)=-3t^2+2t-1$ will take a maximum. Note that $g(t)<0$ for all $t \in \mathbb{R}$ 
So $$ u= {2\over -{2\over 3}} = -3\implies ....$$
