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.

A bit geometry;
1)$x^2+y^2 = 1$, a circle, centre $(0,0)$, $r=1$.
2) Minimum of $C$:
$C:=\dfrac{y+1}{x+y-2}$
(Note: $x+y-2 \not =0$).
$C(x+y-2) = y+1$, or
$Cx +y(C-1) -(2C+1)= 0$, a straight line.
The line touches or intersects the circle 1)
if the distance line-to-origin $\le 1$ (radius).
Distance to $(0,0):$
$d =\dfrac{|2C+1|}{\sqrt{C^2+(C-1)^2}} \le 1.$
$(2C+1)^2 \le C^2 + (C-1)^2;$
$4C^2 +4C +1 \le 2C^2 -2C+1;$
$2C(C+3) \le 0.$
Hence: $-3 \le C \le 0$.
Minimum at $C =-3$.
Used: Line to point distance formula:
http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html
