Getting two tangent circle tangent to their respective vector at a point? Here's my problem

I have two point $A$ and $B$, two vector $\overrightarrow{v}$ and $\overrightarrow{u}$.
I need to find the Center $C_1$ and $C_2$ of the circle $c_1$ and $c_2$.
$c_1$ is tangent to $\overrightarrow{v}$ at $B$.
$c_2$ is tangent to $\overrightarrow{u}$ at $A$.
$c_1$ and $c_2$ are tangent, this tangent point become $M$.
$c_1$ and $c_2$ have the same radius.
Some information we know :
$BC_1$, $C_1M$, $MC_2$, $C_2A$ are all segments of the same size.
$C_1M$, $MC_2$, are parallel.
There can be 4 solution at the same time, the ideal solution is when the shortest path to go to $M$ goes in the same direction as $\overrightarrow{v}$ from $B$ and $\overrightarrow{u}$ from $A$.
 A: I like a geometric solution. Here is the construction I did with The Geometer's Sketchpad.
No coordinate grid, points $A$ and $B$ are defined as you said, and the two vectors are represented as black lines. Let the two normals (red) meet at $C$, and suppose $AC < BC$.

Construct $D$ on $BC$, between $B$ and $C$, such that $BD = AC$. Let $E$ be the midpoint of $CD$. Let $F$ be the midpoint of $AB$. Consider any pair of points $P$ on $AC$ and $Q$ on $BC$, such that $AP = BQ$. In that case, the midpoint of $PQ$ must lie on line $EF$.
Now construct $G$ on $BC$, between $B$ and $C$, such that $CG = CA$. Construct the circle including points $A$, $G$, and $B$. Consider any pair of circles in this quadrant, tangent to the given lines at $A$ and $B$, and tangent to each other (example in blue). The point of tangency must lie on circle $AGB$.
Now two loci have been defined: line $EF$ and circle $AGB$, both shown in green. An intersection of these loci is a point of tangency for one pair of solutions. Intersections are shown for this case at $H$ and $J$. From there it is quick work to construct the circles.
For the other two pairs of solutions, let $D$ fall on the other side of $B$, and let $G$ fall on the other side of $C$.

A: Let the lines perpendicular to $u$ and $v$ at $A$ and $B$ respectively meet at $V$, and set: $VA=a$, $VB=b$, $\angle AVB=\alpha$ (if those lines are parallel then the solution is even simpler and is left to the reader).
Let then $x$ be the signed distance from $A$ to $C_2$ (positive if $A$ is between $V$ and $C_2$, negative otherwise) and construct $C_1$ on line $VB$ such that the signed distance from $B$ to $C_1$ is $\pm x$, using the same conventions for the sign. By the cosine law applied to triangle $VC_1C_2$ we have then:
$$
(a+x)^2+(b\pm x)^2-2(a+x)(b\pm x)\cos\alpha=4x^2.
$$
Solve these two equations to find the possible values of $x$.

EXAMPLE.
For the case illustrated in the question, we have:
$$
a=3,\quad b=4\sqrt2,\quad \alpha=45°.
$$
The quadratic equations above can then be solved for $x$. In the "$+$" case we get:
$$
x=\frac{1}{2} \left(-7+6 \sqrt{2}\pm\sqrt{189-118 \sqrt{2}}\right)\approx\cases{3.09438\\-1.6091},
$$
while in the "$-$" case we get:
$$
x=\frac{1}{2} \left(-7-6 \sqrt{2}
\pm\sqrt{189+118\sqrt{2}}\right)
\approx\cases{1.68971\\-17.175}.
$$
All those results correspond to valid solutions, as one can check with GeoGebra.
