 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 writtenOn the surface, $r_1^2=a^2(5-4\cos \theta )$ and $r_2^2=r_1^2/4$, so $\varphi =0$ 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:Measure $\theta$ from the radius toward the real line, as in the original diagram. The radial field just outside the surface isSince the interior field is zero,Integrating $a\sigma (\theta )$ around the circumference gives total induced charge $-\lambda$ per unit length.