$\left|\frac{z-25}{z-1}\right|=5$
Let $\quad z=x+i y$
$$
\begin{array}{r}
\left|\frac{x+i y-25}{x+i y-1}\right|=5 \\
\frac{|(x-25)+i y|}{|(x-1)+i y|}=5 \\
\frac{\sqrt{(x-25)^2+y^2}}{\sqrt{(x-1)^2+y^2}}=5
\end{array}
$$
Squaring on both sides,
$$
\begin{aligned}
& (x-25)^2+y^2=25\left[(x-1)^2+y^2\right] \\
& \Rightarrow x^2+625-50 x+y^2=25\left[x^2+1-2 x+y^2\right] \\
& \Rightarrow x^2+y^2-50 x+625=25 x^2+25 y^2-50 x+25 \\
& \Rightarrow \quad 24 x^2+24 y^2=600
\end{aligned}
$$

$$
\begin{aligned}
\Rightarrow & & x^2+y^2 & =25 \\
& \therefore & |z| & =\sqrt{x^2+y^2}=\sqrt{25} \\
& & |z| & =5
\end{aligned}
$$
Hence, option (1) is correct.