Given data

The circle is $x^2+y^2=r^2$ with centre $O=(0,0).$

The chord $PQ$ lies on the line $2x+4y=5.$ Dividing by $2$, $x+2y=\frac{5}{2}.$

We are told that the centre of the circumcircle of triangle $OPQ$ lies on the line $x+2y=4.$

Key observation about the circumcenter of $\triangle OPQ$

Since $P,Q$ are points on the circle $x^2+y^2=r^2$, we have $OP=OQ=r.$

Therefore, in triangle $OPQ$, the circumcenter must lie on the perpendicular bisector of $PQ$.

But more importantly, the circumcircle of triangle $OPQ$ is the unique circle passing through $O,P,Q$.

Any circle through $P,Q$ has its centre on the perpendicular bisector of chord $PQ$.

Find the perpendicular bisector condition

The chord line is $x+2y=\frac{5}{2},$ whose normal vector is $(1,2)$.

Hence the perpendicular bisector of chord $PQ$ is a line parallel to the normal direction $(1,2)$, i.e. along the family $y=2x+c$ or equivalently any line perpendicular to $x+2y=\frac52$.

Now, for the circle $x^2+y^2=r^2$, the perpendicular from the centre $O$ to the chord bisects the chord. Since the chord is $x+2y=\frac52,$ its perpendicular bisector passes through $O$ and is therefore along the line $y=2x.$

So the circumcenter of triangle $OPQ$ must lie on $y=2x.$

Use the given condition

The circumcenter also lies on $x+2y=4.$

Solving with $y=2x$: $x+2(2x)=4$ $5x=4$ $x=\frac45,$ $y=2x=\frac85.$

So the circumcenter of $\triangle OPQ$ is $C=\left(\frac45,\frac85\right).$

Use equality of distances from circumcenter

Since $C$ is the circumcenter of triangle $OPQ$, it is equidistant from $O,P,Q$. In particular, $CO=CP=CQ.$

Now,

Use power relation for common chord of two circles

Points $P,Q$ lie on both circles:

• $x^2+y^2=r^2$ (centre $O$)
• circumcircle of $OPQ$ (centre $C$), which also passes through $O$

Their common chord is exactly the line $PQ$, namely $x+2y=\frac52.$

The radical axis of the two circles is obtained by subtracting their equations.

First circle: $x^2+y^2=r^2.$

Second circle has centre $C=\left(\frac45,\frac85\right)$ and radius $\frac{4}{\sqrt5}$, so $\left(x-\frac45\right)^2+\left(y-\frac85\right)^2=\frac{16}{5}.$

Expanding: $x^2+y^2-\frac85x-\frac{16}{5}y=0.$

Subtracting the first circle equation from this gives the radical axis: $-\frac85x-\frac{16}{5}y+r^2=0.$ Multiply by $5$: $-8x-16y+5r^2=0,$ or $x+2y=\frac{5r^2}{8}.$

But this must be the given chord line: $x+2y=\frac52.$

Therefore, $\frac{5r^2}{8}=\frac52.$

Solve: $5r^2=20$ $r^2=4$ $r=2.$

Since $r>\frac{\sqrt5}{2}>0$, we take $r=2.$

Final answer

$\boxed{2}$