A heavy small ring of weight $W$ is free to slide on a smooth surface wire of radius $a$, fixed in a vertical plane. It is attached by a string of length $l$ where
$$2a > l > a\sqrt{2}$$
to a point on the wire in a horizontal line with the centre. Find tension in the string.
Approach :
1. 
Here, If A be the point where string is attached to wire, P be the equilibrium position of string, I get Tension as
$$ \dfrac{W(l^2-2a^2)}{a\sqrt{4a^2-l^2}}$$
2. 
Here, If A be the point where string is attached to wire, P be the equilibrium position of string, I get Tension as
$$ \dfrac{- W(l^2-2a^2)}{a\sqrt{4a^2-l^2}}$$
Clearly, 2nd Approach is wrong as magnitude of tension can't be negative. But why is it wrong ? Why isn't this diagram possible ?
I have verified that with given restriction on $l$, the 2nd diagram should very well be possible. Can anyone point out where am I going wrong ? Thanks!
Considering first the first position
By geometric considerations the angle $\angle PAB = \alpha $ is such that
$$ 2 a \cos\alpha = l $$
Now calling
$$ \vec R = r(\cos(2\alpha),\sin(2\alpha))\\ \vec W = w(0,-1)\\ \vec T =- t(\cos\alpha,\sin\alpha) $$
in equilibrium we have
$$ \vec R + \vec W + \vec T = 0 $$
or
$$ \left\{ \begin{array}{rcl} r \cos (2 \alpha )-t \cos (\alpha )& = & 0 \\ -w-t \sin (\alpha )+r \sin (2 \alpha )& = & 0 \\ 2 a \cos (\alpha )& = & l \\ \end{array} \right. $$
and solving for $r,t,\alpha$ we obtain
$$ \left[ \begin{array}{ccc} t & r & \alpha \\ \frac{\left(2 a^2-l^2\right) w}{a \sqrt{4 a^2-l^2}} & -\frac{l w}{\sqrt{4 a^2-l^2}} & \tan ^{-1}\left(\frac{l}{a},-\frac{\sqrt{4 a^2-l^2}}{a}\right) \\ \frac{\left(l^2-2 a^2\right) w}{a \sqrt{4 a^2-l^2}} & \frac{l w}{\sqrt{4 a^2-l^2}} & \tan ^{-1}\left(\frac{l}{a},\frac{\sqrt{4 a^2-l^2}}{a}\right) \\ \end{array} \right] $$
one of them is discarded.
in the second position, the string can not remain taut.