resistance of a wire connected circular
This is not true. Imagine a circle. It's length is \$L=2\pi r\$. Now assume we are choosing two points on the circle, such that it is dividing the length to two arcs, one of the length \$L_1=2\pi r t\$ and  \$L_2=2\pi r (1-t)\$, where \$t\$ is a general parameter defining the ratio between two lengths (and defined between \$0\$ and \$1\$, inclusive). So, when we connect a multimeter between the two points, we are getting the two lines connected in parallel, so the resistance would be proportional to $$L_1||L_2=\frac{(2\pi r)(2\pi r)t(1-t)}{2\pi r (t+1-t)} = 2\pi r\ t(1-t)$$
So, if the whole wire is \$5\Omega\$, but we want to get \$1.25\Omega\$, which is \$1/4\$ of it, we should solve $$t(1-t)=1/4$$
which is giving \$t=0.5\$. I.e. the points have to be on the opposite sides of the circle's diameter, thus dividing it's length by two.

Update: Actually the above calculation will be true for any closed shape of length \$L\$, not only circle, of course. We can just omit the explicit \$L\$ calculation and write \$L_1=L t\$ and  \$L_2=L(1-t)\$
and thus $$L_1||L_2=\frac{L^2t(1-t)}{L (t+1-t)} = L t(1-t)$$