= Solution
The <method of images> replaces conductors by fictitious charges chosen to reproduce their boundary potentials in the physical domain. The <uniqueness theorem for the electrostatic potential> then identifies the resulting physical exterior field; the fictitious interior field is not the field inside the conductor.
For a source at $d\hat{\mathbf z}$, let $r'=a^2/d$ and $q'=-qa/d$. At a point on $r=a$, the distances $R$ and $R'$ to source and image obey $R'=(a/d)R$. Hence $q/R+q'/R'=0$ everywhere on the sphere. This proves the <image charge for a grounded conducting sphere> construction.
With real charges at $\pm d\hat{\mathbf z}$, superpose two images of charge $-qa/d$ at $\pm a^2\hat{\mathbf z}/d$. For the right-hand charge, take positive force away from the centre. Its <electrostatic force> is
$$
F_0=\frac{q^2}{4\pi\epsilon_0}\left[\frac1{4d^2}-\frac ad\left(\frac1{(d-a^2/d)^2}+\frac1{(d+a^2/d)^2}\right)\right].
$$
Putting $x=d/a$, the <equal-charge balance around a grounded sphere> requires
$$
\boxed{\frac1{4x^2}=x\left[\frac1{(x^2-1)^2}+\frac1{(x^2+1)^2}\right].}
$$
For large $x$, the right side is approximately $2/x^3$, so $x\simeq8$. This is an approximate equilibrium, not an exact cancellation at $x=8$.
To hold the sphere at <electric potential> $V$, add a central image $Q_0=4\pi\epsilon_0aV$. The grounded pair already contributes zero potential on its surface. Therefore
$$
\boxed{F=F_0+\frac{qaV}{d^2}.}
$$
At $d=8a$ this is $F_0(8a)+qV/(64a)$, or approximately $qV/(64a)$ if the small grounded residual is neglected as in the preceding approximation. The charge on the sphere is $Q_0-2qa/d$; fixing its potential is different from independently fixing its net charge.
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