= Solution
In polar coordinates the metric is
$$
ds^2=\frac{4(dr^2+r^2d\theta^2)}{(1-r^2)^2}.
$$
Along a radius, the hyperbolic distance from the origin to Euclidean radius $r$ is therefore
$$
R=\int_0^r\frac{2\,dt}{1-t^2}
=\log\frac{1+r}{1-r},
$$
so $r=\tanh(R/2)$. The Riemannian area element is
$$
dA=\frac{4r}{(1-r^2)^2}\,dr\,d\theta.
$$
Consequently a hyperbolic disc of radius $R$ has area
$$
\begin{aligned}
A(R)
&=\int_0^{2\pi}\int_0^{\tanh(R/2)}
\frac{4r}{(1-r^2)^2}\,dr\,d\theta\\
&=4\pi\sinh^2(R/2)
=2\pi(\cosh R-1).
\end{aligned}
$$
This proves the <area of a hyperbolic disc> formula from the metric.
For area $\pi/2$ we get $\cosh R=5/4$. Hence $\sinh R=3/4$ and $e^R=2$, so $R=\log2$. Tangent equal discs have centers at distance $2R$. Since the centers lie successively on the same radial geodesic,
$$
d(O,c_n)=2nR=n\log4.
$$
If $r_n$ is the Euclidean coordinate of $c_n$, the radial distance formula gives
$$
\frac{1+r_n}{1-r_n}=4^n.
$$
Therefore the <radial chain of equal hyperbolic discs> has
$$
\boxed{r_n=\frac{4^n-1}{4^n+1}}.
$$
\b[No such isometry to the stated upper-half-plane configuration exists for $n\geq3$.] The centers $c_0,\ldots,c_n$ lie on one hyperbolic geodesic, so their images under an isometry would also lie on one geodesic. In the upper-half-plane model, geodesics are vertical lines or semicircles orthogonal to the real axis. Neither type can contain three distinct points of the horizontal line $y=1$, while the centers of the distinct discs $D'_0,\ldots,D'_n$ all lie on that line. This is the <geodesic obstruction for a horizontal chain of hyperbolic discs>.
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