Two charges q1 and q2 are placed at (0, 0, d)

We know that the potential at point P is V = ∑Vi

Where Vi = qi/4πε0, ri is the magnitude of the position vector P

V = 1/4 πε0 ∑ qi/rpi

When (x,y,z) plane is considered, the two charges lie on the z-axis and is separated by 2d. The potential is given as

$\frac{q_{1}}{\sqrt{x^{2}+y^{2}+(z-d)^{2}}}+\frac{q_{2}}{\sqrt{x^{2}+y^{2}+(z+d)^{2}}}=0$

Squaring the equation, we get

x2 + y2 + z2 + [(q1/q2)2+1/(q1/q2)2-1] (2zd)+d2-0

Therefore, the equation of sphere is

x2 + y2 + z2 + 2ux + 2uy + 2wz + g = 0

Centre of the sphere is

$\left(0,0,-d\left[\frac{q_{1}^{2}+q_{2}^{2}}{q_{1}^{2}-q_{2}^{2}}\right]\right)$

And radius is

$r=\frac{2 q_{1} q_{2} d}{q_{1}^{2}-q_{2}^{2}}$