$A,B,C,D\in\mathbb R^2$, $A=(-1,0),D=(1,0)$, $AB, BC,CD$ has constant length, what is the locus of segment $BC$ Denote points $A=(-1,0),D=(1,0)$, and lengths $|AB|=u$, $BC=v$, $CD=w$, where $u,v,w$ are fixed constant. 
Consider $E=$ the locus of the segment $BC$, it form a subset of $\mathbb R^2$ (possibly an empty set). Can we have the explicit expression of its boundary $\partial E$?
 A: The locus of points $B=(x_B, y_B)$  such that have distance $u$ from $A$ is the circumference 
$\mathcal{C}_A=\{(x,y) : (x-x_A)^2+(y-y_A)^2=u^2\}$
The locus of points $C=(x_C,y_C)$ such that have distance $w$ from $D$ is the circumference 
$\mathcal{C}_B=\{(x,y) : (x-x_B)^2+(y-y_B)^2=w^2\}$
So your locus will be the follwing set:
$\Lambda:=\{(x,y) : \exists B\in \mathcal{C}_A , C\in \mathcal{C}_D 
and \lambda\in [0,1]  such that $
$ |BC|=v ,  (x,y)=B+\lambda (C-B)\}$

$B=(x_B, y_B)\in \mathcal{C}_A$  and $C=(x_C, y_C)\in \mathcal{C}_D$  satisfies the condition $|BC|=v$ then we have that 
$(x_B-x_A)^2+(y_B-y_A)^2=u^2$
and 
$(x_C-x_D)^2+(y_C-y_D)^2=w^2$
and 
$(x_B-x_C)^2+(y_B-y_C)^2=v^2$
but 
$(x_B-x_C)^2+(y_B-y_C)^2=(x_B-x_A)^2+$
$+(x_A-x_C)^2+2(x_B-x_A)(x_A-x_C)+$
$+(y_B-y_A)^2+(y_A-y_C)^2+2(y_B-y_A)(y_A-y_C)=$
$=u^2+ (x_A-x_C)^2+ (y_A-y_C)^2 + $
$+2(x_B-x_A)(x_A-x_C) +2(y_B-y_A)(y_A-y_C) =$
$=u^2+(x_A-x_D)^2+(x_D-x_C)^2+$
$+2(x_A-x_D)(x_D-x_C)+ (y_A-y_D)^2+$
$+(y_D-y_C)^2+2(y_A-y_D)(y_D-y_C)+$
$+ 2(x_B-x_A)(x_A-x_C) +2(y_B-y_A)(y_A-y_C) =$
$=u^2+|AD|^2+w^2+2(x_A-x_D)(x_D-x_C)+$
$+2 (y_A-y_D)(y_D-y_C)+ 2(x_B-x_A)(x_A-x_C) +$
$+2(y_B-y_A)(y_A-y_C)=4+u^2+w^2+4(x_C-1)-$
$-2(x_B+1)(x_C+1)-2y_By_C$
So you have that $B$ and $C$ must satisfies the identity 
$2(x_C-x_B)-2x_Bx_C-2y_By_C=v^2-(u^2+w^2)$
