The line $2x-y+1=0$ is tangent to the circle at the point $\left(\mathrm{2,5}\right)$ and the centre of the circle lies on $x-2y=4.$ The square of the radius of the circle is

$2x-y+1=0$ is tangent to the circle
Slope of line $OA=-\frac{1}{2}$
Equation of $OA, \left(y-5\right)=-\frac{1}{2}\left(x-2\right)$
$2y-10=-x+2$
$x+2y=12$
$\therefore$ intersection with $x-2y=4$ will give coordinates of centre

On solving, we get, the centre of the circle is $\left(\mathrm{8,2}\right)$
The radius $\text{OA}=\sqrt{{\left(8-2\right)}^2+{\left(2-5\right)}^2}=\sqrt{36+9}$
$=\sqrt{45}=3\sqrt{5}$ units

Hence, ${\left(OA\right)}^2=45$