How do you find the two arcs to make a path between two points and two tangents? Say I have two points $A$ and $B$ in an $(x,y)$ plane and tangents of the points which are normalised vectors $\mathrm{dir} A$ $\mathrm{dir} B$, how do you find a path made of two arcs that join them together?
Here is a visual of what I mean. Since the images are not clear, the grey vectors from the two tangents are equal in magnitude but the magnitude is unknown.

Note: I am looking for perfect arcs not cubic bezier curves.
 A: Geogebra drawing
Say that you have a line $a$ defined by two points, $A$ and $A'$, and a line $b$ defined by points $B$ and $B'$. In my Geogebra drawing you can adjust directions of lines $a,b$ by moving points $A'$ and $B'$. You want to create a smooth curve APB made of two circular arcs, $AP$ and $PB$, so that arc $AP$ is tangent to line $a$ at point $A$ and arc $PB$ is tangent to line $b$ at point $B$.
First of all, if nothing else is given, there are infinitely many solutions. I will show that exactly one solution can be obtianed by fixing point $D$ on line $a$ so that line $DPE$ is tangent to both arcs at point $P$. 
Introduce $F=a\cap b$. I will discuss only one case 


*

*when point $F$ exists ($a \not\parallel b$)

*point $D$ is between $F$ and $A$

*point $E$ is between $F$ and $B$
Other cases can be discussed in a similar way.

The followin lengths and angle are known:
$p=FA, q=FB, x=FD, \alpha=\angle AFB$
The only problem is to calculate $y=FE$. Note that:
$$DE=DP+EP=DA+EB=(FA-FD)+(FB-FE)=p-x+q-y$$
On the other side:
$$DE^2=FD^2+FE^2-2FD\cdot FE \cos\angle AFB\tag{1}$$
or:
$$(p-x+q-y)^2=x^2+y^2-2xy\cos\alpha$$
$$p^2+q^2-2px+2pq-2py-2xq+2xy-2qy=-2xy\cos\alpha$$
$$p^2+q^2-2px+2pq-2xq=2py-2xy+2qy-2xy\cos\alpha$$
$$(p+q)^2-2x(p+q)=2y(p+q-x(1+\cos\alpha))$$
$$FE=y=\frac{p+q}{2}\ \frac{p+q-2x}{p+q-x(1+\cos\alpha)}$$
Now that you have point $E$, finding point P is easy (because $DP=DA,EP=EB$). 
Center of arc AP can be found by intersecting line perpendicular to $DP$ at $P$ and bisector of angle $PDA$. In a similar way you can find the center of arc $BP$ (construction not shown).
You can play with my Geogebra drawing by moving lines $a$ and $b$ and adjusting position of point $D$. Basically, you can play with blue dots. 

Other cases, for example, when $A$ is between $F$ and $D$ are also interesting. I'll leave detailed analysis up to you:


EDIT: 
In the meantime WDUK added another condition:
$$AD=DP=PE=BE=z$$
In this case, by applying cosine theorem (1) to triangle $FDE$ we get the following equation:
$$(p-z)^2+(q-z)^2-2(p-z)(q-z)\cos\alpha=(2z)^2$$
The solutions are:
$$z_1=\frac{-p-q+(p+q)\cos\alpha+\sqrt{3p^2+2pq+3q^2-8pq\cos\alpha+(p-q)^2\cos^2\alpha}}{2(1+\cos\alpha)}$$
$$z_2=-\frac{p+q-(p+q)\cos\alpha+\sqrt{3p^2+2pq+3q^2-8pq\cos\alpha+(p-q)^2\cos^2\alpha}}{2(1+\cos\alpha)}$$
...and therefore we have two possible solutions, $APB$ and $AQB$:

Geogebra
EDIT 2:
Another example: 
$$FA=p,\ FB=q,\ AD=DP=PE=BE=z$$
In this case, by applying cosine theorem (1) to triangle $FDE$ we get the following equation:
$$(p+z)^2+(q-z)^2-2(p+z)(q-z)\cos\alpha=(2z)^2$$
The positive solution of this quadratic equation is:
$$z=\frac{-p+q+(-p+q)\cos\alpha-\sqrt{3p^2-2pq+3q^2-8pq\cos\alpha+(p+q)^2\cos^2\alpha}}{2(-1+\cos\alpha)}$$
This defines positions of points $D$ and $E$. The rest of the construction is trivial. 

