TITLE: Fundamental group of complement of circle with a line passing through it.
QUESTION [6 upvotes]: Let $X= S^1 \cup y$-axis in the $xy$-plane.  Compute the fundamental group of $\mathbb{R}^3 - X$
Edit:
To clarify, $S^1 =\{(x,y)|x^2 + y^2 = 1\}$ is the unit circle in the $xy$-plane,  so that the $y$-axis and the circle intersect at $(0,\pm 1).$ 
I tried doing this by first "blowing up" the circle and the line to get a homotopy equivalent space.  It seems like when you do this blow up process you are left with a sphere with two holes in it and two line segments attached to the sphere at the endpoints of the line segment.  The holes in the sphere come from $y$-axis and the line segments come from the space between the inside of the circle and the $y$-axis.  Then, forgetting about the line segments for a moment we can project the rest of the space to $\mathbb{R}^2 - D^2$ via stereographic projection.  The line segments are then arcs connecting two points in the plane (the arcs lying outside the plane).  Finally the whole space deformation retracts to a wedge of 3 circles.  Giving that $\pi_1(X) \cong \mathbb{Z}\ast\mathbb{Z}\ast\mathbb{Z}$.
This is a very visual process and I'm not sure my first space after "blowing up" is correct.  Can anyone check this or give me suggestions.  Also:

The space $X$ is not a knot, but can we still use the Wirtinger presentation to compute $\pi_1(X)$ by considering intersections as both over and under crossings?

REPLY [4 votes]: You ask about the Wirtinger presentations, which does extend from complements of knots in $\mathbb R^3$ to complements of graphs.  This is treated in 9.1.9 of Topology and Groupoids for the case of finite graphs:   
"at each vertex $v$  there is a relation $x^{\varepsilon_{1}}_{1}
x^{\varepsilon_{2}}_{2}\dots x^{\varepsilon_{r}}_{r} = 1$, where the
$x_{i} $  are the edges at the vertex and the sign
$\varepsilon_{i}$  is $+1$  if the arrow for $x_{i}$ points towards
the vertex, and $-1$ otherwise;" 
For your example, here is a rough sketch:

so we get generators $a,b,c,d,e$ and relations $a=bcd, c=b^{-1}ed^{-1}$ which presents the free group on the generators $b,e,d$, which corresponds to squirrel's answer.