= Solution
The <method of images> replaces conductors by fictitious charges outside the physical solution domain. Choose images so the real-plus-image potential obeys the specified conductor boundary values. The electrostatic uniqueness theorem then identifies that potential in the physical domain; fields inside the replaced conductor are not physical predictions.
Put the cylinder axis at the origin, real line at $(2a,0)$ and image at $(a/2,0)$. Let $r_1,r_2$ be distances from these lines. The exterior potential can be written
$$
\boxed{\phi=\frac{\lambda}{2\pi\epsilon_0}\ln\frac{2r_2}{r_1}.}
$$
On the surface, $r_1^2=a^2(5-4\cos\theta)$ and $r_2^2=r_1^2/4$, so $\phi=0$ everywhere as required. The additive constant is essential for a grounded cylinder; the two-dimensional potential need not also be zero at infinity.
The physical line feels the image field at separation $3a/2$. Its <force> per unit length is attractive:
$$
\boxed{\frac{\mathbf F}{\mathcal L}=-\frac{\lambda^2}{3\pi\epsilon_0a}\,\hat{\mathbf x}.}
$$
Measure $\theta$ from the radius toward the real line, as in the original diagram. The radial field just outside the surface is
$$
E_r=\frac\lambda{2\pi\epsilon_0}
\left[\frac{a-2a\cos\theta}{a^2(5-4\cos\theta)}
-\frac{a-(a/2)\cos\theta}{(a^2/4)(5-4\cos\theta)}\right].
$$
Since the interior field is zero,
$$
\boxed{\sigma(\theta)=\epsilon_0E_r=-\frac{3\lambda}{2\pi a(5-4\cos\theta)}.}
$$
Integrating $a\sigma(\theta)$ around the circumference gives total induced charge $-\lambda$ per unit length.
\Image[/past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1-gpp2014-cylinder-image.png]
{title=Original calculated exterior equipotentials and induced surface-charge distribution for the grounded cylinder; conductor interior excluded from the field solution}
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