= Solution
A square <circle packing> of equal radii has one circle per cell of side $2r_1$. Its <packing fraction> is therefore
$$
\boxed{\eta_{\rm square}=\frac{\pi r_1^2}{(2r_1)^2}=\frac\pi4.}
$$
For the <alternating-radius square circle packing> shown in the source figure, let $d$ be the nearest unlike-centre separation. A two-circle repeat cell has area $2d^2$ and solid area $\pi(r_1^2+r_2^2)$. Unlike circles require $d\ge r_1+r_2$; like circles on the two sublattices require $\sqrt2d\ge2r_1$ and $\sqrt2d\ge2r_2$. Thus
$$
\boxed{\eta=\frac{\pi(r_1^2+r_2^2)}{2\max\{(r_1+r_2)^2,2r_1^2,2r_2^2\}}.}
$$
All three nonoverlap constraints matter when varying the radius ratio.
Take $r_1\ge r_2$ as in the illustrated arrangement and set $s=r_2/r_1$. If $0\le s\le\sqrt2-1$, the large circles determine the spacing and $\eta=(\pi/4)(1+s^2)$ increases with $s$. If $\sqrt2-1\le s\le1$, unlike contacts determine it and $\eta=(\pi/2)(1+s^2)/(1+s)^2$, whose derivative is proportional to $s-1$ and is nonpositive. The maximum is therefore at the joining point:
$$
\boxed{\frac{r_1}{r_2}=1+\sqrt2,\qquad
\eta_{\max}=\pi\left(1-\frac1{\sqrt2}\right)\simeq0.92015.}
$$
The large circles then touch their large neighbours, while each small circle exactly fills a square interstice. Interchanging the species gives the reciprocal radius ratio. This is the highest <packing fraction> within the specified alternating square arrangement; it is not a claim about every possible unequal-circle packing.
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