[Question]
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## 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:
1. Draw a line that passes through two points.
2. Draw a circle centered at one point such that another point lies on the circle.
3. 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](https://en.wikipedia.org/wiki/Pentagon#Regular_pentagons) (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](https://www.geogebra.org/graphing). 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](/questions/tagged/atomic-code-golf "show questions tagged 'atomic-code-golf'") similar to [proof-golf](/questions/tagged/proof-golf "show questions tagged 'proof-golf'") albeit in a bit of a weird programming language. There is currently [a +22/-0 consensus on the meta](https://codegolf.meta.stackexchange.com/a/16689/56656) that this sort of thing is allowed.
  
[Answer]
# 2 circles, 13 lines, 17 points
[![picture](https://i.stack.imgur.com/9YnhL.png)](https://i.stack.imgur.com/9YnhL.png)
[Try it on GeoGebra](https://www.geogebra.org/geometry/c4ttfqt6)
* 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](http://mathworld.wolfram.com/Polar.html) 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](https://en.wikipedia.org/wiki/Projective_harmonic_conjugate) 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.
[![picture](https://i.stack.imgur.com/3LY56.png)](https://i.stack.imgur.com/3LY56.png)
Now for the fun part. Line FUBAT is [perspective](https://en.wikipedia.org/wiki/Perspectivity) 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](https://en.wikipedia.org/wiki/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.
[Answer]
# ~~7~~ 6 circles, 3 lines
This is a classical pentagon construction, a proof of its correctness can be found [here](https://math.stackexchange.com/a/95614/109451).
[![enter image description here](https://i.stack.imgur.com/Nxqrp.gif)](https://i.stack.imgur.com/Nxqrp.gif)
[Answer]
# 4 circles, 7 lines
Since it has been beaten I thought I would just post my original solution to the problem. This solution is modified from the method given by Dixon in *Mathographics*, a proof of correctness for that method can be found [here](https://math.stackexchange.com/a/95614/276060).
* Draw \$\mathrm{Circle}(A,B)\$
* Draw \$\overline{AB}\$
* Mark the intersection of \$\mathrm{Circle}(A,B)\$ and \$\overline{AB}\$ as \$C\$
* Draw \$\mathrm{Circle}(B,C)\$
* Draw \$\mathrm{Circle}(C,B)\$
* Mark the intersection of \$\mathrm{Circle}(C,B)\$ and \$\mathrm{Circle}(B,C)\$ as \$D\$
* Mark the intersection of \$\mathrm{Circle}(C,B)\$ and \$\overline{AB}\$ as \$E\$
* Draw \$\overline{DC}\$
* Mark the intersection of \$\mathrm{Circle}(C,B)\$ and \$\overline{DC}\$ as \$F\$
* Mark the intersection of \$\mathrm{Circle}(C,B)\$ and \$\mathrm{Circle}(B,C)\$ as \$G\$
* Draw \$\overline{BG}\$
* Mark the intersection of \$\overline{BG}\$ and \$\overline{EF}\$ as \$H\$
* Draw \$\overline{HC}\$
* Mark the intersection of \$\overline{HC}\$ and \$\mathrm{Circle}(C,B)\$ as \$I\$
* Draw \$\overline{IA}\$
* Mark the intersection of \$\overline{IA}\$ and \$\mathrm{Circle}(A,B)\$ as \$J\$
* Draw \$\mathrm{Cirlce}(I,J)\$
* Mark the intersection of \$\mathrm{Circle}(I,J)\$ and \$\overline{HC}\$ as \$L\$
* Mark the intersections of \$\mathrm{Circle}(I,J)\$ and \$\mathrm{Circle}(C,B)\$ as \$M\$ and \$K\$.
* Draw \$\overline{ML}\$
* Draw \$\overline{KL}\$
* Mark the intersection of \$\mathrm{Circle}(C,B)\$ and \$\overline{ML}\$ as \$N\$
* Mark the intersection of \$\mathrm{Circle}(C,B)\$ and \$\overline{HC}\$ as \$O\$
* Mark the intersection of \$\mathrm{Circle}(C,B)\$ and \$\overline{KL}\$ as \$P\$
\$MKPON\$ is a regular pentagon.
[![Drawing](https://i.stack.imgur.com/xs63F.png)](https://i.stack.imgur.com/xs63F.png)
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