Given that
$\left|\frac{\mathrm{z}-25}{\mathrm{z}-1}\right|=5 \Rightarrow|\mathrm{Z}-25|=5|\mathrm{Z}-1|$
Let $Z=x+i y$, then
$\begin{array}{l}
|x+i y-25|=5|x+i y-1| \\
\Rightarrow|(x-25)+i y|=5 \mid(x-1+i y \mid
\end{array}$
Squaring both sides, we get
$\begin{array}{l}
(\mathrm{x}-25)^{2}+\mathrm{y}^{2}=25\left\{(\mathrm{x}-1)^{2}+\mathrm{y}^{2}\right\} \\
\Rightarrow \mathrm{x}^{2}-50 \mathrm{x}+625+\mathrm{y}^{2} \\
\quad=25 \mathrm{x}^{2}-50 \mathrm{x}+25+25 \mathrm{y}^{2} \\
\Rightarrow 24 \mathrm{x}^{2}+24 \mathrm{y}^{2}-600=0 \\
\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}-25=0 \\
\Rightarrow|\mathrm{x}+\mathrm{iy}|^{2}=25 \Rightarrow|\mathrm{Z}|^{2}=5^{2} \\
\Rightarrow|\mathrm{Z}|=5
\end{array}$