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.

Just for a variation, using Lagrange’s method:
$$
f(a,b,t)=\frac{b+1}{a+b-2}-t(a^2+b^2-1)
$$
Then
\begin{align}
\frac{\partial f}{\partial a}&=-\frac{b+1}{(a+b-2)^2}-2at \\[6px]
\frac{\partial f}{\partial b}&=\frac{a-3}{(a+b-2)^2}-2bt
\end{align}
If these equal $0$, then
$$
-\frac{b+1}{a(a+b-2)^2}=\frac{a-3}{b(a+b-2)^2}
$$
so that $-b^2-b=a^2-3a$, that gives $3a-b=1$ and from $a^2+b^2=1$ we derive $a=0$ or $a=3/5$.
The critical points are thus $(0,-1)$ and $(3/5,4/5)$. We have
$$
f(0,-1,0)=\frac{2}{3},\quad
f(3/5,4/5,0)=-3,\quad
$$
This also shows the maximum.
