$\begin{aligned}
& x^2-10 x+y^2=0 \\
& x-10 x+25+y^2=25 \Rightarrow \text { centre }=(5,0) \text { and } r=5
\end{aligned}$
$\mathrm{y}=2 \mathrm{x}$ is a chord of given circle.
Point of intersection of chord and circle is
$\mathrm{x}^2-10 \mathrm{x}+25+4 \mathrm{x}^2=25 \quad \Rightarrow \mathrm{y}=0,4$
Thus end points of the chord are $(0,0)$ and $(2,4)$
Mid point of the chord $=\left(\frac{2}{2}, \frac{4}{2}\right)=(1,2)$ and
length of chord $=\sqrt{(2)^2+(4)^2}=\sqrt{20}$ is the diameter of required circle.

Hence equation of required circle is $(x-1)+(y-2)=\left(\frac{\sqrt{20}}{2}\right)^2$ i.e. $x^2+y^2-2 x-4 y=0$