Two point charges -Q and +Q / 3 are placed in the x y -plane at the...

Let $P$$\left(h, k\right)$ be be a general point with potential to be zero.

${V}_{1} + {V}_{2} = 0$ (as potential given $= 0$)

$\dfrac{K\left(- Q\right)}{\sqrt{{h}^{2} + {k}^{2}}} + \dfrac{K\left(\dfrac{Q}{\sqrt{3}}\right)}{\sqrt{{\left(h - 2\right)}^{2} + {k}^{2}}} = 0$

$\dfrac{1}{\sqrt{3}\sqrt{{\left(h - 2\right)}^{2} + {k}^{2}}} = \dfrac{1}{\sqrt{{h}^{2} + {k}^{2}}}$

$3\left[{\left(h - 2\right)}^{2} + {k}^{2}\right] = {h}^{2} + {k}^{2}$

$3\left({h}^{2} - 4h + 4 + {k}^{2}\right) = {h}^{2} + {k}^{2}$

$2{h}^{2} + 2{k}^{2} - 12h + 12 = 0$

${h}^{2} + {k}^{2} - 6h + 6 = 0$

\left({h}^{2} - 6h + 6\right) + {k}^{2} = 0$</p><p>\left({h}^{2} - 6h + 9\right) + {k}^{2} = 3$

${\left(h - 3\right)}^{2} + {k}^{2} = 3$

Standard equation of circle is given by

${\left(x - {x}_{1}\right)}^{2} + {\left(y - {y}_{1}\right)}^{2} = {R}^{2}$

$∴ {x}_{1} = 3, {y}_{1} = 0$ and $R = \sqrt{3}$

Hence, $R = \sqrt{3}m$

and value of $b = 3m$