Envelope of family of curves $x(u,v)=\cos^2(u)\cos(v)+\cos(u)\sin(u)\sin(v)$, $y(u,v)=\cos^2(u)\sin(v)-\cos (u)\sin(u)\cos(v)$ How to generally find singular solution or envelope of a two parameter family of curves $ x(u,v),y(u,v) $ in the plane?
The parametric equations
$$x(u,v) = \cos^2 (u) \cos (v) + \cos (u) \sin (u) \sin (v),\\
 y(u,v) = \cos^2 (u) \sin (v) - \cos (u) \sin (u) \cos (v),\\
 (0 < v <  2 \pi), (0 < u < \pi ),$$
represent rotating circles of unit diameter passing through the origin and rotating about the origin. How do we obtain their envelope $ x^2+y^2=1 ? $
EDIT1:
The envelope is $x^2+y^2==1 $ as shown below.
EDIT 2:
We can wlog write $v = c $ for any rotated position.
$$x(u,c) = \cos^2 (u) \cos (c) + \cos (u) \sin (u) \sin (c),\\
 y(u,c) = \cos^2 (u) \sin (c) - \cos (u) \sin (u) \cos (c).\\ $$

I think that I know the C-discriminant method with one parameter, but would like to know how to extend it to two parameters.  
Can it be extended to 3D space to find a surface envelope?
 A: So if I understood you correctly, we have  the curves $\gamma_v(u):(0, \pi)\to\mathbb R^2$, given by:
$$\gamma_v(u)=\begin{pmatrix}x_v(u)\\y_v(u)\end{pmatrix} = 
\begin{pmatrix}\cos^2 (u) \cos (v) + \cos (u) \sin (u) \sin (v)\\
  \cos^2 (u) \sin (v) - \cos (u) \sin (u) \cos (v)\end{pmatrix}$$
and we are looking for the envelope of the family $\{\gamma_v(u)\mid v\in [0,2\pi)\}$.
I believe that you already found that each $\gamma_v(u)$ describes the circle $$\left( x-\frac12 \cos(v)\right)^2+\left( y-\frac12\sin(v)\right)^2=\frac14.$$
We now convert to the implicit form
$$F(u,v)=\left( x-\frac12 \cos(v)\right)^2+\left( y-\frac12\sin(v)\right)^2-\frac14=0.$$ 
Note that $(0,0)$ lies on $F(u,v)=0$, for every $v$.
We can now find our envelope by solving:
$$\begin{cases} F(u,v)=0\\ \\ \dfrac{\partial F(u,v)}{\partial v}=0\end{cases}$$
You can easily check that $\dfrac{\partial F}{\partial v}=x\sin(v)-y\cos(v).$ So we find that points lie on our envelope, if and only if, it is an intersection point of the circle $F(u,v)=0$ and the line $x\sin(v)-y\cos(v)=0$, for some $v$.
Since we can easily see that $\left(\frac12 \cos(v), \frac12 \sin(v)\right)$ (the centre of $F(u,v)=0$) always lies on the line $x\sin(v)-y\cos(v)=0$, we infer geometrically (see picture) that the envelope is:
$$(0,0)\cup \left\{ (x,y)\mid x^2+y^2=1\right\}.$$
Note how the origin should be included in the envelope.


$\color{red}{\text{The circles $F(u,v)=0$ (red)}}$;
  The line $x\sin(v)-y\cos(v)=0$;
  $\color{blue}{\text{The envelope (blue)}}$. 

The same result can be derived using algebraic methods, but this is a lot more writing.
