Observation 6.1. Let $(G, S, T, g, k)$ be an instance of Planar Disjoint Paths. Let $R'$ be a Steiner tree. Then, $|V_1(R')| = 2k$ and $|V_{\ge 3}(R')| \le 2k - 1$.
Before we proceed to the next step, we claim that every vertex of $H$ is, in fact, “close” to the vertex set of $R_1$. For this purpose, we need the following proposition by Jansen et al. [23].
Proposition 6.1 (Proposition 2.1 in [23zia]). Let $G$ be a plane graph and with disjoint subsets $X, Y \subseteq V(G)$ such that $G[X]$ and $G[Y]$ are connected graphs and $\mathrm{rdist}_G(X, Y) = d \ge 2$. For any $r \in {0, 1, \dots, d-1}$, there is a cycle $C$ in $G$ such that all vertices $u \in V(C)$ satisfy $\mathrm{rdist}_G(X, {u}) = r$, and such that $V(C)$ separates $X$ and $Y$ in $G$.
Additionally, we need the following simple observation.
Observation 6.2. Let $G$ be a triangulated plane graph. Then, for any pair of vertices $u, v \in V(G)$, $\mathrm{dist}_G(u, v) = \mathrm{rdist}_G(u, v)$.
Proof. Let $\mathrm{rdist}G(u, v) = t$, and consider a sequence of vertices $u = x_1, x_2, \dots, x{t+1} = v$ that witnesses this fact—then, every two consecutive vertices in this sequence have a common face. Since $G$ is triangulated, we have that ${x_i, x_{i+1}} \in E(G)$ for every two consecutive vertices $x_i, x_{i+1}$, $1 \le i \le t$. Hence, $x_1, x_2, \dots, x_{t+1}$ is a walk from $u$ to $v$ in $G$ with $t$ edges, and therefore $\mathrm{dist}_G(u, v) \le \mathrm{rdist}G(u, v)$. Conversely, let $\mathrm{dist}G(u, v) = \ell$; then, there is a path with $\ell$ edges from $u$ to $v$ in $G$, which gives us a sequence of vertices $u = y_1, y_2, \dots, y{\ell+1} = v$ where each pair of consecutive vertices forms an edge in $G$. Since $G$ is planar, each such pair of consecutive vertices $y_i, y{i+1}$, $1 \le i \le \ell$, must have a common face. Therefore, $\mathrm{rdist}_G(u, v) \le \mathrm{dist}_G(u, v)$. $\square$
It is easy to see that Observation 6.2 is not true for general plane graph. However, this observation will be useful for us because the graph $H$, where we construct the backbone Steiner tree, is triangulated. We now present the promised claim, whose proof is based on Proposition 6.1, Observation 6.2 and the absence of long sequences of $S \cup T$-free concentric cycles in good instances. Here, recall that $c$ is the fixed constant in Corollary 4.1. We remark that, for the sake of clarity, throughout the paper we denote some natural numbers whose value depends on $k$ by notations of the form $\alpha_{\text{subscript}}(k)$ where the subscript of $\alpha$ hints at the use of the value.
Lemma 6.1. Let $(G, S, T, g, k)$ be a good instance of Planar Disjoint Paths. Let $R'$ be a Steiner tree. For every vertex $v \in V(H)$, it holds that $\mathrm{dist}H(v, V(R')) \le \alpha{\mathrm{dist}}(k) := 4 \cdot 2^{ck}$.
Proof. Suppose, by way of contradiction, that $\mathrm{dist}H(v^*, V(R')) > \alpha{\mathrm{dist}}(k)$ for some vertex $v^* \in V(H)$. Since $H$ is the (enriched) radial completion of $G$, it is triangulated. By Observation 6.2, $\mathrm{rdist}H(u,v) = \mathrm{dist}H(u,v)$ for any pair of vertices $u,v \in V(H)$. Thus, $\mathrm{rdist}H(v^*, V(R')) > \alpha{\mathrm{dist}}(k)$. By Proposition 6.1, for any $r \in {0, 1, \dots, \alpha{\mathrm{dist}}(k)}$, there is a cycle $C_r$ in $H$ such that all vertices $u \in V(C_r)$ satisfy $\mathrm{rdist}H(v^*, u) = \mathrm{dist}H(v^*, u) = r$, and such that $V(C_r)$ separates ${v^}$ and $V(R')$ in $H$. In particular, these cycles must be pairwise vertex-disjoint, and each one of them contains either (i) $v^$ in its interior (including the boundary) and $V(R')$ in its exterior (including the boundary), or (ii) $v^*$ in its exterior (including the boundary) and $V(R')$ in its interior (including the boundary). We claim that only case (i) is possible. Indeed, suppose by way of contradiction that $C_i$, for some $r \in {0, 1, \dots, \alpha{\mathrm{dist}}(k)}$, contains $v^*$ in its exterior and $V(R')$ in its interior. Because the outer face of $H$ contains a terminal $t^* \in T$ and $t^* \in V(R')$, we derive that $t^* \in V(C_i)$. Thus, $\mathrm{rdist}H(v^*, t^*) = i \le \alpha{\mathrm{dist}}(k)$. However, because $t^* \in V(R')$, this is a contradiction to the supposition that $\mathrm{dist}H(v^*, V(R')) > \alpha{\mathrm{dist}}(k)$. Thus, our claim holds true. From this, we conclude that $C = (C_0, C_1, \dots, C{\alpha{\mathrm{dist}}(k)})$ is a $V(R')$-free sequence of concentric cycles in $H$. Since $S \cup T \subseteq V(R')$, it is also $S \cup T$-free.
Consider some odd integer $r \in {1, 2, \dots, \alpha_{\text{dist}(k)}}$. Note that every vertex $u \in V(C_r)$ that does not belong to $V(G)$ lies in some face $f$ of $G$, and that the two neighbors of $u$ in $C_r$ must belong to the boundary of $f$ (by the definition of radial completion). Moreover, each of the