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 $.
Working on achille hui idea in his comment above, if $\theta_i$ is the angle connecting side $i$ to side $(i+1)$, where side 6 is side 1, then,
$\tan\left(\dfrac{\theta_i}{2}\right) = \dfrac{R}{x_{i+1}} \tag{1}$
where it is assumed that moving counter clockwise (for example) and segmenting the side $s_i$ to $s_i = x_i + x_{i+1} $ where the point of separation is the tangency point with the inscribed circle.
It is easy to construct a linear system to relate the $5$ $x_i$'s to the $5$ side lengths.
Now, we know that the sum of the interior angles in a pentagon is $3(180^\circ) = 540^\circ$, therefore the sum of half the angles is $270^\circ$
Further, the tangent of the sum of $5$ angles is given by (see this Wikipedia page)
$ \tan(\phi_1 + \phi_2 + \phi_3 + \phi_4 + \phi_5) = \dfrac{ \sum t_i - \sum t_i t_j t_k + \sum t_i t_j t_k t_l t_m }{ 1 - \sum t_i t_j + \sum t_i t_j t_k t_l } $
where $t_i = \tan(\theta_i) $
Since $270^\circ$ has an infinite tangent, we want the denominator in the above expression to $0$. The denominator is a function of $R^2$ because the $x_i's$ are known. Hence we end up with a quadratic equation in $R^2$ which is trivial to solve.
Once $R$ is found, we can compute the center of the inscribed circle, and compute all the internal angles, and construct the pentagon and the inscribed circle.