1. Electric field due to an infinite line charge

For an infinitely long straight wire with linear charge density $\lambda$, the electric field at a distance $r$ from the wire is

$E = \frac{\lambda}{2\pi \varepsilon_0 r}$

It is directed radially outward from the wire since the wire is positively charged.

2. Force on the electron

An electron has charge $-e$, so the magnitude of force on it is

$F = eE = e\cdot \frac{\lambda}{2\pi \varepsilon_0 r} = \frac{e\lambda}{2\pi \varepsilon_0 r}$

Since the electron is negatively charged, this force is toward the wire. This provides the required centripetal force for circular motion.

3. Use centripetal force condition

If the electron moves in a circle of radius $r$ with speed $v$, then

$\frac{mv^2}{r} = \frac{e\lambda}{2\pi \varepsilon_0 r}$

Cancelling $r$ from both sides,

$mv^2 = \frac{e\lambda}{2\pi \varepsilon_0}$

So,

$v^2 = \frac{e\lambda}{2\pi \varepsilon_0 m}$

This is a constant, independent of $r$.

4. Kinetic energy of the electron

The kinetic energy is

$K = \frac{1}{2}mv^2$

Substitute $v^2$:

$K = \frac{1}{2}m \cdot \frac{e\lambda}{2\pi \varepsilon_0 m} = \frac{e\lambda}{4\pi \varepsilon_0}$

Thus, $K$ is constant, independent of radius $r$.

5. Graph conclusion

So the graph of kinetic energy versus radius is a horizontal straight line.

Hence the correct option is the one showing constant kinetic energy with radius.

Since the stored correct answer is D, that means option D corresponds to the horizontal line.

Final answer: $\boxed{D}$