Combining my second guess with CarpetPython's answer (green line is what to plow):
$4.87 text{km}$

Second guess:
Ok, I think I can get it down to $5.196 text{km}$

If the circle is inscribed in a hexagon, first draw 3 of the spokes (1.155 km each). From the tips of the spokes, imagine it creates the equilateral triangle. From the center of each face, draw a line to the closest corner. Those lines are 1/2 the length of the first set of spokes. So, $4.5 times 1.155 = 5.196 text{km}$ The red lines below are the plow lines.

First guess:
If you plow an X across the circle, with leg lengths of $sqrt{2} text{km}$, you can get your plowing down to $5.656 text{km}$ (or imagine that the circle in in a square and you plow both diagonals).

If the tunnel is on the tangent of the circle (say, at the very top) it will barely hit each of the ends of the X.

I think the 4.83 solution is this:

5.217 km

I used inscribed (green) and circumscribed (black) squares to help. The original circle is shown in blue. The plow paths are shown in red. The black square has a side of length $2$ and the green square has a side length of $sqrt2$. The plow path lengths are as follows:

A = $sqrt2 = 1.414$

B = $sqrt2div2 = 0.707$

C = $sqrt{sqrt2div2} = 0.841$

Total Plow Length $ = 5.217$

Even if my math is right, this is still not the optimal answer according to OP's source. I hope that it at least throws out some new ideas, though.

4.8736 km

I have modified my answer when I observed that a central blocking element would result in less plowing. I wrote a short program to try a range of angles (as my calculus and geometry are too rusty) to find the best angle of 30 degrees.

EDIT: Note that JonTheMon posted his 4.87 answer first while I was still programming, so he gets the credit for first with this answer.

 A  Plow Distance
 19 4.91250640267
 20 4.90597796428
 21 4.90001860583
 22 4.89463591311
 23 4.88983859219
 24 4.88563652529
 25 4.88204083336
 26 4.87906394597
 27 4.87671967936
 28 4.87502332331
 29 4.87399173788
 30 4.87364346116   <- best
 31 4.87399882899
 32 4.8750801084
 33 4.87691164606

First Solution: 5.14 km

First plow the southern half of the circumference, then from each end plow 1 km north (making a U shape). The total distance will be pi+2 or 5.14 km.

NOTE: Not a complete answer. The purpose of this answer is to possibly offer an idea that someone else can turn into a formal answer.

Things to consider:

1. Experiment with the length of the lines
2. Experiment with the best angle of the lines, relative to the tangent of the circle at that point
3. Experiment with the number of lines used.
4. Experiment with adding a small circle to the center, if needed.

4.8264 km

Plough a $120°$ section of the circle, extending both ends in a straight line by a distance of $sqrt{1/3}$ km to meet the vertices of a hexagon enclosing the circle. Two more lines of length $1$ km and $sqrt{1/3}$ km are ploughed as shown to cover paths through the other half of the circle. The total length ploughed is equal to:

$$frac{2pi}{3} + 3 times sqrt{frac13} + 1 = 4.82644591 text{km}$$

(This puzzle is related to the opaque square problem discussed in Professor Stewart's Casebook of Mathematical Mysteries)

4.8205 km (or 4.8189 km with slightly more work).

Let $ABCDE$ be a circumscribed regular pentagon around the circle, let $M$ be the point of tangency between segment $BC$ and the circle, and let $N$ be the point of tangency between segment $DE$ and the circle. Let $X$ be the midpoint of line $BE$.

Plow the arc of the circle connecting $M,N$, and plow the line segments $AX, BM, NE$. The total length is

$frac{4pi}{5} + |AX| + 2|BM| = frac{4pi}{5} + frac{3sqrt{5}-5}{2} + 2sqrt{5-2sqrt{5}} = 4.82046...$.

---

To get 4.8189...

Instead of using a regular pentagon use a slightly irregular circumscribed pentagon. If the pentagon is symmetric around $AX$, with the angle at $C$ equal to $pi-x$ and the angle at $B$ equal to $frac{pi-y}{2}$, then the total length comes out to

$2x + cos(x+y)sec(y)-sec(x+2y) + 2tan(y)$,

and plugging in $x approx 1.259557, y approx 0.6432556$ we get $4.8189...$.

---

We can also get a lower bound on the total length. First note that we have to plow at least one point with each possible $x$ coordinate between $1$ and $-1$, so the length of the projection of our plow path onto the $x$-axis is at least $2$. Thus the original length is at least 2 km.

Slightly more sophisticatedly, we can combine the fact that the projection onto the $x$-axis has length at least $2$ with the fact that the projection of the length onto the $y$-axis has at least $2$ to see that the total length is at least $2sqrt{2}$ km: the average of the two projections of any straight line of length $l$ onto the $x$ and $y$ axes is at most $frac{l}{sqrt{2}}$, and the total of these averages over all the line segments we plow must be at least $2$.

Taking this to its logical conclusion, if we try averaging the projection of a straight line segment onto a uniformly random direction, we get $frac{2l}{pi}$, so the total length plowed must be at least $pi$ km.

4.8179 km

I have made a little improvement on @squeamish ossifrage answer where it just requires a little optimization on it, resulting;

with the minimization of the equation.

First of all, using squeamish ossifrage's idea, I have found every angle is equal to each other except the arc one, so I asked myself why it has to be that way;

The red line shows how we are supposed to plow, which is found earlier but not with the optimal angle. So I suspected that the length of the arc and the rest is not optimal even though it seems they are.

Then I defined $|GJ|$ and $|HL|$ in terms of $a$ as below (let me know if you want to show me how I found these equations for each);

$|HL|=2cdot sec{a} cdot sin^{2}a$

To find $|HL|$, I put another illustration below:

As you see, we already know that $|AI|=|AH|=sec{a}$ and the angle $measuredangle {HAI}=2a$ so;

$|AL|=sec{a} cdot cos{2a}$

so,

$|HL|=|AH|-|AL|=sec{a}-sec{a} cdot cos{2a}=sec{a}(1-cos{2a})=2cdot sec{a} cdot sin^{2}a$

and lastly,

$|GJ|=4sin^{2}a$

To show that, I put a zoomed picture on that area;

1. $measuredangle {FAM}=3a$ as a result $measuredangle {MFA}$ becomes $frac{pi}{2}-3a$.
2. $|AF|=|AG|$ and $measuredangle {FAG}=2a$ so $measuredangle {GFA}=frac{pi}{2}-a$
3. $measuredangle {GFM}=measuredangle {GFA}-measuredangle {MFA}=2a$

Since we know $|GF|=2tan{a}$ then;

$|GJ|=2tan{a}cdot sin{2a}=4sin^2{a}$

Moreover, we now that the tangent lines values are $tana$ already. As a result, we define the total length of the plow;

$P(a)=2pi-8a+2tan{a}+2cdot sec{a} cdot sin^{2}a+4sin^{2}a$ $P(a)=2pi-8a+2tan{a}+2sin^{2}aleft (2+ sec{a} right )$

so we need to minimize this function; as a result mentioned at the beginning, the angle $a$ becomes;

Note: The illustration is made with the result angle $a$.