Construct a pentagon avoiding compass use Rules

You will start with only two elements: Points \$A\$ and \$B\$ such that \$A \neq B\$.  These points occupy a plane that is infinite in all directions.

At any step in the process you may do any of the three following actions:


Draw a line that passes through two points.
Draw a circle centered at one point such that another point lies on the circle.
Add a new point where two objects (lines and circles) intersect.


Your goal is to create 5 points such that they form the vertices of a regular pentagon (a convex polygon with 5 sides equal in length) using as few circles as possible.  You may, of course, have other points but 5 of them must for a regular pentagon.  You do not have to draw the edges of the pentagon for your scoring.

Scoring

When comparing two answers the one that draws fewer circles is better.  In the case of a tie in circles the answer that draws the fewest lines is better.  In the case of a tie in both circles and lines the answer that adds the fewest points is better.

Anti-Rules

While the rules list is exhaustive and details everything you can do, this list is not. Just because I don't say you can't do something does not mean you can.


You cannot create "arbitrary" objects.  Some constructions you will find will do things like add a point at an "arbitrary" location and work from there.  You cannot add new points at locations other than intersections.
You cannot copy a radius.  Some constructions will involve taking a compass setting it to a radius between two points and then picking it up and drawing a circle elsewhere.  You cannot do this.
You cannot perform limiting processes.  All constructions must take a finite number of steps.  It is not good enough to approach the answer asymptotically.
You cannot draw an arc or part of a circle in order to avoid counting it as a circle in your scoring.  If you want to visually use arcs when showing or explaining your answer because they take up less space go ahead but they count as a circle for scoring.


Tools

You can think through the problem on GeoGebra.  Just go over to the shapes tab.  The three rules are equivalent to the point, line and circle with center tools.

Burden of Proof

This is standard but I would like to reiterate.  If there is a question as to whether a particular answer is valid the burden of proof is on the answerer to show that their answer is valid rather than the public to show that the answer is not.

What is this doing on my Code-Golf site?!

This is a form of atomic-code-golf similar to proof-golf albeit in a bit of a weird programming language. There is currently a +22/-0 consensus on the meta that this sort of thing is allowed.
2 circles, 13 lines, 17 points



Try it on GeoGebra


Let circle(A, B) intersect circle(B, A) at C and D.
Let AB intersect circle(A, B) again at E.
Let AB intersect circle(B, A) again at F.
Let AD intersect circle(A, B) again at G.
Let AD intersect CF at H.
Let BG intersect DF at I.
Let HI intersect circle(A, B) at J and K.
Let BG intersect EJ at L.
Let BJ intersect EG at M.
Let BG intersect EK at N.
Let BK intersect EG at O.
Let LM intersect circle(A, B) at P and S.
Let NO intersect circle(A, B) at Q and R.


Then EPQRS is a regular pentagon.

Why it works

Let BE intersect GJ at T, and let BE intersect GK at U.  The complete quadrilateral BEGJ shows that T is the polar of LM, which is the intersection of the tangents at P and S.  Similarly, the complete quadrilateral BEGK shows that U is the polar of NO, which is the intersection of the tangents at Q and R.

Let FG intersect HI at V.  The diagonals DV and GI of the complete quadrilateral DGVI intersect FH at harmonic conjugates with respect to F and H; since the first is at ∞, the second is the midpoint C of FH, which is to say that C, D, V are collinear.

Let CG intersect HI at W.



Now for the fun part.  Line FUBAT is perspective about G to line VKIHJ, which is perspective about D to circle CKDGJ, which is perspective about C to line HKVWJ, which is perspective about G to line AUF∞T.  Composing these four perspecitivities yields a projectivity FUBAT ⌅ AUF∞T.  Since a one-dimensional projectivity is determined by three points, T and U are determined as the two fixed points of FBA ⌅ AF∞.

Assigning coordinates with A = 0, B = −1, F = −2, this projectivity is defined by x ↦ 4/x + 2, and its fixed points T = 1 + √5 = sec(2π/5) and U = 1 − √5 = −sec(2π/10), exactly as required to make EPQRS a regular pentagon.