Path connectedness of the set $\{(x,y):(x+1)^{2}+y^{2}\leq 1\}\cup\{(x,y):y=x\sin(\frac{1}{x}),x>0\}$

Let $A$ be the following subset of $\mathbb{R}^{2}$:$$A=\{(x,y):(x+1)^{2}+y^{2}\leq 1\}\cup\{(x,y):y=x\sin(\frac{1}{x}),x>0\}$$ Then

$1.$ $A$ is connected.

$2.$ $A$ is compact.

$3.$ $A$ is path connected.

$4.$ $A$ is bounded.

It is clear that $A$ is unbounded as the graph of $x\sin(\frac{1}{x})$, so option $2$nd and $4$th are not correct. Again $A$ is connected by using the fact that closure of connected set is connected. But i am confused about path connectedness of $A$. Please help me about path connectedness of $A.$ Thanking you.

Name $$A_1=\{(x,y):(x+1)^{2}+y^{2}\leq 1\}$$ and $$A_2=\{(x,y):y=x\sin(\frac{1}{x}),x>0\}$$

You have $A_1 \cap \overline{A_2} = \{(0,0)\}$ where $\overline{A_2} =A_2 \cup \{(0,0)\}$ and $A = A_1 \cup \overline{A_2}$.

$\overline{A_2}$ is path connected. That can be proved considering the path $$\gamma(t)=\begin{cases} (0,0) & t = 0\\ (at, at \sin \frac{1}{at}) & t \in (0,1] \end{cases}$$ for $a >0$.

$A_1$ is convex therefore path connected. Finally $A$ is path connected as a union of two path connected sets with non empty intersection.