The pentagon has been surrounded!

From the equation of family of lines:

$L \equiv \left(1+2\lambda\right)y - \left(2+\lambda\right)x +\left(1-\lambda\right)=0 \\ ~~~ \equiv \left(y-2x+1\right) + \lambda\left(2y-x-1\right)=0$

The intersection point of this family of lines is $\left(1,1\right)$ .

The figure above (not to scale) shows the diagram for this problem. Here $O\left(x,y\right)$ is the centre of circumcircle of pentagon and the two non-overlapping triangles having no common side are shown in yellow. The length of each of the diagonals is $L=\sin36^{\circ}$ and the radius of circumcircle is $R$ . The magnitude of angle subtended at the centre by each side of the pentagon is $\dfrac{360^{\circ}}{5}=72^{\circ}$ as shown above.

From the figure, we have:

$L=2\left(R\sin72^{\circ}\right)$ $\Longrightarrow \sin36^{\circ}=2R\sin72^{\circ}$ $\Longrightarrow R=\dfrac{\sin36^{\circ}}{2\sin72^{\circ}}=\dfrac{2\sin18^{\circ}\cos18^{\circ}}{2\cos18^{\circ}}=\sin18^{\circ}=\cos72^{\circ}$

Using distance formula, we obtain the locus of centre of circumcircle $O\left(x,y\right)$ as:

$\text{Locus} \equiv {\left(x-1\right)}^{2} + {\left(y-1\right)}^{2} = R^{2} \\ \qquad \equiv x^{2}+y^{2}-2x-2y+2=\cos^{2}72^{\circ} \\ \qquad \equiv \boxed {x^{2}+y^{2}-2x-2y+1+\sin^{2}72^{\circ}=0} ~~~~~~~~~~ \small \color{#3D99F6}{\text{As,} ~2-\cos^{2}72^{\circ} = 1+\left(1-\cos^{2}72^{\circ}\right) = 1+\sin^{2}72^{\circ}}$