25. As shown in Figure 1, in the upper half-plane of the coordinate plane, there are several circles forming a set. These circles are all tangent to the $x$-axis. The outermost layer (the $L_{0}$ layer) consists of two externally tangent circles with radii $70^{2}$ and $73^{2}$. On $\bigcup_{j=0}^{k-1} L_{j}(k \geqslant 1)$, for every two adjacent circles, a new circle is exactly externally tangent to both of these circles. According to this rule, the $L_{k}$ layer has $2^{k-1}$ circles.

Let $S=\bigcup_{j=0}^{6} L_{j}$, and for each circle $C$, its radius is denoted as $r(C)$. Then $\sum_{C \in s} \frac{1}{\sqrt{r(C)}}$ is ( ).
(A) $\frac{286}{35}$
(B) $\frac{583}{70}$
(C) $\frac{715}{73}$
(D) $\frac{143}{14}$
(E) $\frac{1573}{146}$
25. D.

Let the larger circle in layer $L_{0}$ be $X$, with radius $x$, and the smaller circle be $Y$, with radius $y$. The circle in layer $L_{1}$ is $Z$, with radius $z$.

Let the distance between the two points where circle $X$ and $Z$ touch the $x$-axis be $a$, and the distance between the two points where circle $Y$ and $Z$ touch the $x$-axis be $b$. Then the distance between the two points where circle $X$ and $Y$ touch the $x$-axis is $a+b$.
According to the problem, we have
$$
\begin{array}{l}
(x-z)^{2}+a^{2}=(x+z)^{2} \Rightarrow a=2 \sqrt{x z}, \\
(y-z)^{2}+b^{2}=(y+z)^{2} \Rightarrow b=2 \sqrt{y z}, \\
(x-y)^{2}+(a+b)^{2}=(x+y)^{2} \\
\Rightarrow a+b=2 \sqrt{x y} .
\end{array}
$$

Thus, $2 \sqrt{x z}+2 \sqrt{y z}=2 \sqrt{x y}$
$$
\Rightarrow \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=\frac{1}{\sqrt{z}} \text {. }
$$

Let $S(n)$ denote the sum of the reciprocals of the square roots of the radii of the circles in the $n$-th layer. Then
$$
\begin{array}{l}
S(0)=S(1)=\frac{1}{\sqrt{z}}=\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=\frac{1}{73}+\frac{1}{70}, \\
S(2)=S(0)+2 S(1)=3 S(0), \\
S(3)=2 S(2)+2 S(1)+S(0)=9 S(0), \\
S(4)=2 S(3)+2 S(2)+2 S(1)+S(0) \\
=27 S(0), \\
S(5) \\
=2 S(4)+2 S(3)+2 S(2)+2 S(1)+S(0) \\
=81 S(0), \\
S(6)=2 S(5)+2 S(4)+2 S(3)+ \\
=243 S(0) . \\
\text { Therefore, } \sum_{C \in S} \frac{1}{\sqrt{r(C)}}=\sum_{k=0}^{6} S(k) \\
=(1+1+3+9+27+81+243) S(0) \\
=365 S(0)=365\left(\frac{1}{73}+\frac{1}{70}\right)=\frac{143}{14} .
\end{array}
$$