Let D be the Poincare' disk its natural hyperbolic metric and with at least 1 marked point on $\partial D$. Suppose I cut an hyperbolic circle of radius $r$ away from it, then I get a Riemann surface with a boundary $\partial C$, which is not a geodesic boundary. However, it is possible to find a Riemannian metric on $D\setminus C$ so that $\partial C$ is now geodetic.
My question is, it is possible to do so without changing the metric on $D\setminus C$, or at least only by a total rescaling?
My intuition is that, if I rescale the whole metric, I can make the geodesic curvature of $\partial C$ small at will, so it should be possible.
I made a tentative proof which is as follows: Quite obviosuly, we need to add an annulus. Any annulus can be written as $A_M = \{ z \in \mathbb{C} |\ R < |z| < 1\}$, $M$ being its modulus given by $$M = \ln{1/R}.$$ It is well known that we can find a generator $\pi_1(A_M)$ which is a geodetic - with respect to the hyperbolic metric of the annulus - of length equal to $\pi/M$ (or something very similar, not really relevant). We can of course scale $A_M$ by a multiplication $z \to z/r$, without changing its modulus, this might change the geodetic representative of $\pi_1(A_M)$ to a new one which we call $\gamma$.
Now I glue the outer boundary of the annulus to $\partial C$. This can be done in many ways, because of the twists, but I can use the marked points on $\partial D$ to define this gluing uniquely: we can define marked points on $\partial C$ by projecting them through the old center of $C$.
Now, I argue that the surface I get carries an hyperbolic metric for which the punctures on $\partial D$ are half-cusps and $\gamma$ is a geodetic of length uniquely identified by the (former) radius of $C$, r. Indeed, the surface I got is locally isometric to $D$ or to $A_M$ by construction, and this should end the proof.
Am I correct?
EDIT:
I forgot the uniqueness part which is at follows: Suppose two annuli exists with different moduli and the property of having exterior radius r, then we can glue them together by this common circle and we get a surface with two geodetic boundaries of different lengths, which is impossible.
RE-EDIT: I made a lot of confusion in asking the question, so I restate it completely:
Basically: Is the moduli space of the Poincare' disk with $n$ marked points on the boundary, and a marked circle inside, equivalent to the moduli space of Riemann surfaces with two geodesic boundaries, and $n$ marked points on one of them?
I thought I could cut away the circle and then uniquely determine, in terms of its radius, center and relative locations to the markings, an annulus to attach to it so that I get a Riemann surface with two geodetic boundaries.
Also,I see now that my proof is wrong because I attach an infinite length component of the annulus to the former circle, and that does not make any sense.
