The circle is $x^2+y^2=r^2$ with center at $O(0,0)$ and radius $r$.

The given line is $y-2x=3 \\Rightarrow -2x+y-3=0.$

The length of the chord cut by a line at distance $d$ from the center in a circle of radius $r$ is $L=2\sqrt{r^2-d^2}.$ Given that the chord length is $r$, we have $2\sqrt{r^2-d^2}=r.$

First find the perpendicular distance of the center $(0,0)$ from the line $-2x+y-3=0$: $d=\frac{| -2(0)+1(0)-3 |}{\sqrt{(-2)^2+1^2}}=\frac{3}{\sqrt{5}}.$ So, $d^2=\frac{9}{5}.$

Now use the chord-length condition: $2\sqrt{r^2-\frac{9}{5}}=r.$ Squaring both sides, $4\left(r^2-\frac{9}{5}\right)=r^2.$ $4r^2-\frac{36}{5}=r^2.$ $3r^2=\frac{36}{5}.$ $r^2=\frac{12}{5}.$

Compare with the options: $r^2=\frac{12}{5}$ which is Option C.