[proofplan] We compare the circuit families of the two matroids under the primal-dual edge bijection. The key geometric fact is that circuits in the plane dual are exactly the duals of bonds in the primal graph: ordinary dual cycles and dual loops correspond respectively to non-bridge bonds and bridge bonds in the primal graph. Since bonds are precisely the cocircuits of a finite connected graphic matroid, and cocircuits of $M(G)$ are circuits of $M(G)^*$ by the definition of matroid duality, the circuit families agree. [/proofplan]

[step:Fix the primal-dual edge notation] Let $E := E(G)$ be the finite edge set of $G$. The plane dual $G^\dagger$ has vertex set equal to the face set of the fixed plane embedding of $G$, and for each edge $e \in E$ there is a unique dual edge $\iota(e) \in E(G^\dagger)$ joining the two faces incident to $e$, with the convention that if the two incident faces coincide then $\iota(e)$ is a loop. Thus $\iota: E \to E(G^\dagger)$ is a bijection. For a subset $A \subset E$, define $A^\dagger := \iota(A) \subset E(G^\dagger)$. We identify subsets of $E$ with their images under $\iota$ only when explicitly stated; all comparisons below are made through the map $A \mapsto A^\dagger$. [/step]

[step:Show that dual cycles are exactly primal bonds]We prove that a subset $B \subset E$ is a bond of $G$ if and only if $B^\dagger$ is a circuit of the graphic matroid $M(G^\dagger)$. In this proof, a circuit of a graph means a minimal nonempty dependent edge set in its graphic matroid: it is either the edge set of an ordinary simple cycle or a single loop. This convention is needed because plane duals of bridges are loops. Here a cut means a set of the form $\delta_G(S) := \{e \in E(G) : e \text{ has one endpoint in } S \text{ and one endpoint in } V(G) \setminus S\}$ for a nonempty proper subset $S \subset V(G)$, and a bond means an inclusion-minimal nonempty cut. For an edge set $F \subset E(G)$, the notation $G - F$ means the graph obtained from $G$ by deleting the edges in $F$ and retaining all vertices. For a subset $S \subset V(G)$, the induced subgraph $G[S]$ means the graph with vertex set $S$ and edge set consisting of all edges of $G$ whose endpoints both lie in $S$. [claim:Cycle-cut correspondence for a fixed plane embedding] For every subset $B \subset E$, the set $B^\dagger$ is a circuit of the graphic matroid $M(G^\dagger)$ if and only if $B$ is a bond of $G$. [/claim] [proof] We invoke the standard planar bond-cycle correspondence theorem for finite connected plane multigraphs: under the edge-dual bijection of a fixed plane embedding, an edge set is an inclusion-minimal nonempty cut in the primal graph if and only if its dual edge set is a circuit of the dual graphic matroid. The hypotheses of this external theorem apply here because $G$ is connected, finite, and equipped with a fixed plane embedding, and $G^\dagger$ is the plane dual built from that embedding. The theorem includes loops: a primal bridge corresponds to a dual loop, and that loop is a one-edge circuit of $M(G^\dagger)$. To record what this cited theorem asserts geometrically, recall its two ingredients. First, if $B^\dagger$ is an ordinary simple cycle in the embedded dual, its image is a simple closed curve on the sphere meeting the primal graph exactly in the crossings with the edges of $B$. By the Jordan curve theorem, this curve has two complementary components. The primal edges crossing the curve are precisely the edges with endpoints on opposite sides, so $B = \delta_G(S)$ for the set $S$ of primal vertices on one side. The minimality of this cut is part of the planar bond-cycle correspondence: if a proper nonempty subset of $B$ were already a cut, then its dual edges would contain a nonempty proper closed subwalk of the simple dual cycle, contradicting minimality of the circuit in the graphic matroid. The one-edge case is the same statement with a dual loop and a primal bridge. Second, if $B$ is a bond, then deleting $B$ separates $G$ into exactly two connected components with vertex sets $S$ and $T := V(G) \setminus S$; otherwise a component of either side would define a smaller nonempty cut contained in $B$. In the sphere, the common frontier of sufficiently small regular neighbourhoods of the embedded connected subgraphs $G[S]$ and $G[T]$ crosses exactly the edges of $B$. The planar bond-cycle correspondence says that this frontier reads as a minimal closed dual walk. Minimality excludes a repeated dual vertex before return to the starting vertex, because a repeated vertex would split off a smaller nonempty closed dual walk and hence, by the Jordan curve theorem applied to that closed subwalk after removing immediate backtracking, a smaller nonempty primal cut contained in $B$. Thus $B^\dagger$ is a circuit of $M(G^\dagger)$. Therefore $B^\dagger$ is a circuit of $M(G^\dagger)$ if and only if $B$ is a bond of $G$. [/proof][/step]

[guided]We want to translate a planar statement into a matroid statement, so the first task is to identify which primal edge sets become circuits in the dual graph. A cut is a set $\delta_G(S) := \{e \in E(G) : e \text{ has one endpoint in } S \text{ and one endpoint in } V(G) \setminus S\}$ for a nonempty proper subset $S \subset V(G)$, and a bond is an inclusion-minimal nonempty cut. For an edge set $F \subset E(G)$, the notation $G - F$ means that we delete the edges in $F$ and keep all vertices. In a multigraph, a circuit of the graphic matroid means an ordinary simple cycle edge set or a single loop; this is the right convention here because a bridge in $G$ becomes a loop in $G^\dagger$. The precise planar input is the standard planar bond-cycle correspondence theorem for finite connected plane multigraphs, used here as an external topological graph theorem. It says that, under the edge-dual bijection of a fixed plane embedding, an edge set $B \subset E(G)$ is a bond of the primal graph if and only if $B^\dagger$ is a circuit of the dual graphic matroid. Its hypotheses are satisfied: $G$ is connected, the embedding is fixed, the graph is finite, and $G^\dagger$ is the dual built from the faces of that embedding. Why is this the correct topological statement? In one direction, an ordinary simple dual cycle is drawn as a simple closed curve on the sphere. The Jordan curve theorem applies to this curve and gives exactly two complementary components. The construction of the plane dual makes the curve meet the primal graph exactly at the crossings of the dual edges with their corresponding primal edges. Therefore the crossed primal edges are exactly the edges with endpoints on opposite sides of the curve, so they form a cut $\delta_G(S)$ for the set $S$ of primal vertices on one side. The minimality assertion is not a separate hand-waving step: it is part of the planar bond-cycle correspondence. A proper nonempty subcut would dualise to a proper nonempty closed subwalk contained in the dual cycle, contradicting that the dual edge set is a circuit. If the dual circuit is a loop, the same statement says that the crossed primal edge is a bridge, which is exactly a one-edge bond. In the other direction, suppose $B$ is a bond. Then $G - B$ has exactly two connected components with vertex sets $S$ and $T := V(G) \setminus S$. If one side were disconnected, a connected component of that side would determine a smaller nonempty cut contained in $B$, contradicting the minimality of the bond. Now take sufficiently small regular neighbourhoods of the two embedded connected subgraphs $G[S]$ and $G[T]$ on the sphere. Their common frontier crosses precisely the edges of $B$, and it crosses each of them once. The cited planar bond-cycle correspondence theorem supplies the required topological verification: the frontier read through adjacent faces gives a minimal closed dual walk. If the closed dual walk repeated a dual vertex before returning to its starting point, the repeated-vertex segment would contain a smaller nonempty closed dual walk; after cancelling immediate reversals, the Jordan curve theorem applied to a simple closed subcurve would produce a smaller nonempty primal cut contained in $B$. That contradicts that $B$ is a bond. Hence the dual edge set is a circuit of $M(G^\dagger)$. Thus $B^\dagger$ is a circuit of $M(G^\dagger)$ exactly when $B$ is a bond of $G$, which is the planar translation needed for the matroid argument.[/guided]

[step:Identify primal bonds with circuits of the dual matroid] For a connected graph $G$, the circuits of the graphic matroid $M(G)$ are the edge sets of cycles in $G$, and the cocircuits of $M(G)$ are exactly the bonds of $G$. Indeed, a cocircuit is a minimal subset of $E(G)$ whose deletion increases the number of connected components of $G$, and those minimal disconnecting edge sets are precisely the bonds. By the definition of the dual matroid $M(G)^*$, the circuits of $M(G)^*$ are the cocircuits of $M(G)$. Therefore, for every subset $B \subset E$, $B$ is a circuit of $M(G)^*$ if and only if $B$ is a bond of $G$. [/step]

[step:Compare the circuit families and obtain the matroid isomorphism] Combining the previous two steps, for every subset $C \subset E$, $C$ is a circuit of $M(G)^*$ if and only if $C$ is a bond of $G$. The bond-cycle correspondence gives that $C$ is a bond of $G$ if and only if $C^\dagger$ is a circuit of $M(G^\dagger)$. Hence $C$ is a circuit of $M(G)^*$ if and only if $\iota(C)$ is a circuit of $M(G^\dagger)$. A matroid is determined by its circuit family, since its independent sets are exactly the subsets containing no circuit. Thus the bijection $\iota: E(G) \to E(G^\dagger)$ preserves and reflects circuits, and therefore induces an isomorphism of matroids $M(G)^* \cong M(G^\dagger)$. This is the desired planar duality for graphic matroids. [/step]