= Solution
Substituting the wire potential from part (b) into the flux formula from part (a) gives <Neumann's mutual-inductance formula>
$$
L_{12}
=\frac{\Phi_{12}}{I_2}
=\frac{\mu_0}{4\pi}
\oint_{C_1}\oint_{C_2}
\frac{dx_1\mathbin{\cdot}dx_2}{|x_1-x_2|}.
$$
Interchanging the two curves proves $L_{12}=L_{21}$.
Parametrize the coaxial circles by
$$
x_1=(a\cos\phi,a\sin\phi,0),
\qquad
x_2=(b\cos\psi,b\sin\psi,c).
$$
Then, with $\theta=\phi-\psi$ and $R=\sqrt{a^2+b^2+c^2}$,
$$
dx_1\mathbin{\cdot}dx_2
=ab\cos\theta\,d\phi\,d\psi,
\qquad
|x_1-x_2|=R\sqrt{1-q\cos\theta},
$$
where $q=2ab/R^2$. One angular integration contributes $2\pi$, so
$$
L_{12}
=\frac{\mu_0ab}{2R}
\int_0^{2\pi}\frac{\cos\theta\,d\theta}
{\sqrt{1-q\cos\theta}}.
$$
Since $ab=qR^2/2$, this is
$$
\boxed{
L_{12}=\frac{\mu_0R}{4}f(q)
=\frac{\mu_0}{4}\sqrt{a^2+b^2+c^2}\,f(q)}.
$$
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