If $\frac{|z-2|}{|z-3|}=2$  represents a circle, then its radius is equal to:

Given $\frac{\left|z-2\right|}{\left|z-3\right|}=2$ , let $z=x+iy$  then

$\Rightarrow$     $\sqrt{{\left(x-2\right)}^2+y^2}=2\sqrt{{\left(x-3\right)}^2+y^2}$

$\Rightarrow$        ${\left(x-2\right)}^2+y^2=4\left[{\left(x-3\right)}^2+y^2\right]$

$\Rightarrow$     $x^2+y^2+4-4x=4x^2+4y^2+36-24x$

$\Rightarrow$            $3x^2+3y^2-20x+32=0$

or         $x^2+y^2-\frac{20}{3}x+\frac{32}{3}=0$                                   …..(i)

we know that, standard equation of circle,

         $x^2+y^2+2gx+2fy+c=0$                                ……(ii)

Comparison of Eq. (i) from Eq. (ii)

$\Rightarrow$        $2g=-\frac{20}{3}\Rightarrow g=-\frac{10}{3},f=0,c=\frac{32}{3}$

Hence, Radius $=\sqrt{g^2+f^2-c}=\sqrt{\frac{100}{9}-\frac{32}{3}}=\sqrt{\frac{4}{9}}=\frac{2}{3}$