geometrypolygons
You're given the consecutive lengths of the sides of an irregular pentagon, and you want to circumscribe this pentagon (whose interior angles are unknown yet) about a circle of unknown radius.
Is there a known procedure to do this ?
The image below is one that I created in Microsoft Excel using VBA code that implements a Newton-Raphson iteration in the the five interior angles of the pentagon, using consecutive side lengths of $3, 6, 6, 3, 2 $.
For a $n$ sided polygon, you need all the angles in order and $n-2$ consecutive side lengths in order to construct the polygon. So, you need the lengths of sides $B,C$ or $E,B$ or $D,E$ to construct your polygon.
The best way to find out the length of the remaining side is by drawing diagonals and applying triangle laws (sine or cosine rule).
Consider the (very badly drawn) pentagon. It is not drawn to scale, but you get the idea. Here are the steps you will take to find out the lengths of $D,E$.
1.Find out length of $X$ using cosine rule in $\Delta ABX$.
2.Knowing $\angle a,X,A,B$, find out $\angle e,\angle b$ using sine rule.
3.$\angle c = \angle\mbox{(between B,C)} -\angle b$. So, $\angle c$ is known.
4.Repeat the whole procedure for $\Delta CXY$. Find out $Y,\angle d, \angle f$.
5.$\angle g, \angle h$ are easily calculated now.
6.$\angle i$ is known. Apply sine rule in $\Delta DEY$ to find out $D,E$-the two unknown sides.
Each pentagon consists of an equilateral triangle and a rhombus, see figure (and the Note at the end) 
To obtain a desired tiling, you can consider the length of each side equal $1.$
The center of the configuration is in $0.$ Starting with two points (here $0$ and $1$), any further point is obtained by a rotation of a point we already have. With the use of complex numbers has a rotation a simple formula $$z'-z_0=e^{i\varphi}(z-z_0),$$ where $z_0$ is ordinate of the center of the rotation, $z$ is that of the point you want to rotate, and $z'$ is the obtained image of $z.$ Complex ordinates of some points are enclosed.
Note added: there is a typo in the picture, the most right point is obtained by a clockwise rotation (negative angle), the right term is $e^{-i10\pi/18}.$
Best Answer
Here the solution for the inradius is made more explicit.
We use the diagram below to define sides $s_k$ and vertex angles $2\alpha_k$, which are bisected to give half-angles $\alpha_k$ (the latter appear in the drawing as $a_k$ because I cannot access Greek font in that app).
Drawing the radius to each side abd applying SOH CAH TOA to the resulting right triangles leads to the following:
$s_1=r(\cot\alpha_1+\cot\alpha_2)$
$s_2=r(\cot\alpha_2+\cot\alpha_3)$
$s_3=r(\cot\alpha_3+\cot\alpha_4)$
$s_4=r(\cot\alpha_4+\cot\alpha_5)$
$s_5=r(\cot\alpha_5+\cot\alpha_1)$
We solve these for the cotangents in terms of the inradius $r$ and the sides getting
$\cot\alpha_k=\Delta_k/r\tag{1}$
$\Delta_k=\frac12(s_k-s_{k+1}+s_{k+2}-s_{k+3}+s_{k+4})\tag{2}$
where $s_n$ for $n\ge6$ is reduced by using the wraparound criterion $s_n\equiv s_{n-5}$.
Before we go further we note carefully that the half-angles $\alpha_k$ are acute, meaning the cotangents all must be positive. Therefore $\Delta_k$ in Eq. 2 must also be positive and we have a constraint on the sides of the pentagon:
A tangential pentagon must have each pair of nonadjacent sides summing to less than half the total perimeter.
For instance, if the consecutive sides are specified as $1,2,3,4,5$ then the perimeter is $15$ and the nonadjacent sides $3$ and $5$ sum to $8\not<15/2$. Ergo, no go. Fortunately, the OP's pentagon with sides $3,6,6,3,2$ has perimeter $20$ and all nonadjacent side pairs summing to $\le9<20/2$ (there is no objection to the adjacent long sides adding up to $12$). So we may proceed.
We have the $\alpha_k$, but need the inradius $r$ to evaluate them. So we apply an implicit definition from the known sum of the vertices for a convex pentagon:
$\alpha_1+\alpha_2+\alpha_3+\alpha_4+\alpha_5=270°\tag{3a}$
We also have a root where the sides form a pentagram, with the circle inside the inner pentagon. In this solution, which always exists in tandem with the convex one, the vertex angles sum to $180°$ and therefore the half-angles sum to $90°$:
$\alpha_1+\alpha_2+\alpha_3+\alpha_4+\alpha_5=90°\tag{3b}$
We now combine these equations by taking the cotangent of the half-angle sum, which will be zero:
$\cot(\alpha_1+\alpha_2+\alpha_3+\alpha_4+\alpha_5)=0\tag{4}$
The cotangent of a sum of two arguments is given by
$\cos(u+v)=\dfrac{\cot u \cot v -1}{\cot u + \cot v},$
and by iterating on this formuka we derive the formuka for the cotangentbof the sum of the five half-angles:
$\cos(\alpha_1+\alpha_2+\alpha_3+\alpha_4+\alpha_5)=\dfrac{\sum[\cot\alpha_i]-\sum[(\cot\alpha_i)(\cot\alpha_j)(\cot\alpha_k)]+(\cot\alpha_1)(\cot\alpha_2)(\cot\alpha_3)(\cot\alpha_4)(\cot\alpha_5)}{\sum[(\cot\alpha_i)(\cot\alpha_j)(\cot\alpha_k)(\cot\alpha_l)]-\sum[(\cot\alpha_i)(\cot\alpha_j)]+1},$
where each summation is taken over all nonrepeating permutations of the indices from $1$ to $5$.
Then for the cotangent of the sum to be zero (Eq. 4) the numerator must be zero. Plugging that in and using Eq. 2 to express the cotangents in terms of $r$ leads to the quartic equation
$\sum[\Delta_i]r^4-\sum[(\Delta_i)(\Delta_j)(\Delta_k)]r^2+(\Delta_1)(\Delta_2)(\Delta_3)(\Delta_4)(\Delta_5)=0.\tag{5}$
We now solve Eq. 5 as a quadratic for $r^2$, extract positive squareoots to obtain the inradii $r_+$ (the larger root representing a convex pentagon) and $r_-$ (the smaller root representing a pentagram). Then Eq. 1 will guve the $\alpha$ values which are doubled to obtain the vertex angles.
Let's do this for the OP's pentagin with sides $3,6,6,3,2$. We already saw that the side-length constraint is satisfied, so the $\Delta$ values should all be positive. Eq. 2 duly gives $\Delta_k=1,2,4,2,1$. From these values Eq. 5 is rendered:
$10r^4-64r^2+16=0\implies 5r^4-32r^2+8=0.$
Then using the quadratic firmula followed by taking square roots yields
$r_+=\sqrt{\dfrac{16+6\sqrt6}{5}}\approx2.4778$
$r_-=\sqrt{\dfrac{16-6\sqrt6}{5}}\approx0.5105$
The larger root agrees with the size of the circle in the OP's graph, which touches the $x$-axis and tops out just short of $y=5$.
We apply Eq. 1 to compute the vertex angles for both roots, obtaining:
$r_+\approx2.4778,2\alpha_1=2\alpha_5\approx136°02'36'',2\alpha_2=2\alpha_4\approx102°10'51'',2\alpha_3\approx63°33'07''$
$r_-\approx0.5105,2\alpha_1=2\alpha_5\approx54°05'20'',2\alpha_2=2\alpha_4\approx28°38'17'',2\alpha_3\approx14°32'46''.$