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.

From
$$
\frac{b+1}{a+b-2}= u\Rightarrow L\to a = 2-b\frac{b+1}{u}
$$
So $L$ should be tangent to $a^2+b^2=1\;$ then substituting we have the condition
$$
(2b^2-4b+3)u^2+(4+2b-2b^2)u+(b+1)^2 = 0
$$
and solving for $u$
$$
u = \frac{(b+1)^2}{b^2-b\pm\sqrt{(1-b) (b+1)^3}-2}
$$
but tangency implies on $\sqrt{(1-b) (b+1)^3}=0\;$ hence the solutions for tangency are $b = \pm 1$ etc.
