Why do we lose information from constraint equations at branch cut points?

Consider a pendulum doing vertical circular motion and having velocity with magnitude $v$, mass $m$ and length of string $l$ starting from horizontal (inextensible rope, the given velocity is at horizontal position).

Using force balance at top most point (and considering tension is 0),

$$ frac{mv_{tngt}^2}{r} = mg$$

Which gives, $$ |v_{tngt}| = sqrt{rg}.$$

Now by the consideration that at top most point the vertical velocity must turn to zero,

$$ v_{tngt} = |v_{tngt}| vec{i}.$$

Now, consider another way of doing this using constraint equations, the pendulum moves in a path of circle centered at origin:

$$ x^2 +y^2 = l^2.$$

Differentiating this constraint equation with respect to time,

$$ x v_x + y v_y = 0.$$

Now, this equation relates the components of velocity in cartesian coordinates but the problem is that if you put the $ (x,y)$ of top most point which is $ (0,l)$ there is no useful information to be derive:

$$ 0 v_x + l v_y = 0 .$$

And similarly this happens for the point $ (l,0) , (0,-l) , (-l,0)$ at these points we can’t derive any useful information of motion about motion using constraint. I can’t understand why at at these points these equations do not give us new information.. as in , is there a deeper understanding to it than it being an algebraic result?

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After some deep thought, I noted that all these points are where the branch cuts of the circle equation occurs if we write the constraint as an explicit function of $x$ or $ y$.

For example, if we wrote the isolate the circle equation for $x$ and square rooted, we would get two equations,

$$ x = sqrt{r^2 – y^2}$$

$$ x= – sqrt{r^2 -y^2}.$$

This two equations we got from square root can be thought of the circle being split into two half discs along the vertical line $ x=0$ with each equation of the above denoting one half disc. This vertical line cuts the circle at $ (0,l) $ and $ (0,-l)$ so I am guessing the resolution has to do with something about where the explicit function’s definition shifts but I’m not sure.