Two point charges $-Q$ and $+Q / \sqrt{3}$ are placed in the $x y$-plane at the origin $(0,0)$ and a point…

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

$V_1+V_2=0$   (as potential given $=0$)

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

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

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

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

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

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

$\left(h^2-6h+6\right)+k^2=0$

$\left(h^2-6h+9\right)+k^2=3$

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

Standard equation of circle is given by

${\left(x-x_1\right)}^2+{\left(y-y_1\right)}^2=R^2$

$\therefore x_1=3, y_1=0$ and $R=\sqrt{3}$

Hence, $R=\sqrt{3} \mathrm{m}$

and value of $b=3 \mathrm{m}$