If I am looking at a collection $\mathcal{C}$ of circles $\{C_1,...,C_n\}$ all of which have some radii $\{r_1,...,r_n\}$ where $r_i\in\mathbb{R}^{+}$ for each $i \in[n]$. In $\mathcal{C}$, all the circles are pairwise intersecting i.e. each circle intersects another circle. We do not allow touchings (i.e. the circles only touch at one single point). Note that we also do not allow faces in these arrangements that are bounded by two arcs and two intersection points.

We define a $3$-cell in this collection of circles as a cell bounded by three arcs and three intersection points of the circles. Moreover, an empty $3$-cell is a $3$-cell such that no circle passes through it.

I am trying to prove or disprove the following: each circle $C_i\in\mathcal{C}$ bounds two empty $3$-cells so that one of them is inside $C_i$ and one of them on the outside of $C_i$.

I have the some of these following ideas: fix $C_i$ and give it a certain direction and split the intersection points of $C_i$ with other circles into two sets: $\{p_1,...,p_{n-1}\}$ being the first intersections of $C_i$ with $C_j$ for $i\neq j$ and $\{q_1,...,q_{n-1}\}$ being the set of the second such intersections.

Now, it seems like any set of three circles creates exactly $7$ $3$-cells, however, in the entire $\mathcal{C}$ these are not necessarily empty, and this is the problem. Therefore as a whole, there are the total of $7\cdot \binom{n}{3}$ $3$-cells.

Another thing worth noting looking at is how the circles behave on the inner and the outer side of the fixed $C_i$ based on the order of their intersection with $C_i$, but I am not entirely sure how to go from here.

Perhaps

If I am looking at a collection $\mathcal{C}$ of circles $\{C_1,...,C_n\}$ all of which have some radii $\{r_1,...,r_n\}$ where $r_i\in\mathbb{R}^{+}$ for each $i \in[n]$. In $\mathcal{C}$, all the circles are pairwise intersecting i.e. each circle intersects another circle. We do not allow touchings (i.e. the circles only touch at one single point). Note that we also do not allow faces in these arrangements that are bounded by two arcs and two intersection points.

We define a $3$-cell in this collection of circles as a cell bounded by three arcs and three intersection points of the circles. Moreover, an empty $3$-cell is a $3$-cell such that no circle passes through it.

I am trying to prove or disprove the following: each circle $C_i\in\mathcal{C}$ bounds at least two empty $3$-cells so that one of them is inside $C_i$ and one of them on the outside of $C_i$.

I have the some of these following ideas: fix $C_i$ and give it a certain direction and split the intersection points of $C_i$ with other circles into two sets: $\{p_1,...,p_{n-1}\}$ being the first intersections of $C_i$ with $C_j$ for $i\neq j$ and $\{q_1,...,q_{n-1}\}$ being the set of the second such intersections.

Now, it seems like any set of three circles creates exactly $7$ $3$-cells, however, in the entire $\mathcal{C}$ these are not necessarily empty, and this is the problem. Therefore as a whole, there are the total of $7\cdot \binom{n}{3}$ $3$-cells.

Another thing worth noting looking at is how the circles behave on the inner and the outer side of the fixed $C_i$ based on the order of their intersection with $C_i$, but I am not entirely sure how to go from here.

Perhaps