Let $X=\{(x,y,z)\in\mathbb{R}^3|x^2+y^2+z^2=9\}\cup\{(x,y,0)\in\mathbb{R}^3|(x-4)^2+y^2=1\}\subset\mathbb{R}^3$
Find the fundamental group $\pi_1(X,p_0)$ where $p_0=(3,0,0)$
Possible solution: I'm trying to use van Kampen principle. We have a plane that is tangent to a circle. I believe that I need to divide my space into two open sets.
Let $U = \{(x,y,z)\in\mathbb{R}^3|x^2+y^2+z^2<10\}\cap X$ And $V$ is a sphere with some radius (lets say $3$) based in the center of the circle, intersected with $X$.
I believe the next steps are these (But I don't know how to prove all of them):
$1$. $U\cup V=X$ (this is trivial)
$2$. $U\cap V$ is path-connected with a trivial fundamental group
$3$. The fundamental group of $U$ is trivial (this is easy)
$4$. The fundamental group of $V$ is $\mathbb{Z}$ (this is known)
$5$. Thus $\pi_1(X,p_0) = \mathbb{Z}$. I believe this is possible because of point $2$ but I am not sure why.
Please help.