\[\begin{array}{*{35}{l}}

arg(\left| \left( z\text{ }\text{ }-2 \right)\text{ }/\text{ }\left( z-\text{ }\text{ }3 \right) \right|\text{ })=\text{ }2  \\

Substituting\text{ }z\text{ }=\text{ }x\text{ }+\text{ }iy,\text{ }we\text{ }get  \\

\Rightarrow \left| \left( x\text{ }+\text{ }iy\text{ }\text{ }-2 \right)\text{ }/\text{ }\left( x\text{ }+\text{ }iy\text{ }\text{ }-3 \right) \right|\text{ }=\text{ }2  \\

\Rightarrow ~\left| x\text{ }\text{ }-2\text{ }+\text{ }iy \right|\text{ }=\text{ }2\text{ }\left| x\text{ }\text{ }-3\text{ }+\text{ }iy \right|  \\

\Rightarrow \surd \left( {{\left( x\text{ }\text{ }-2 \right)}^{2}}~+\text{ }{{y}^{2}} \right)\text{ }=\text{ }2\surd \left( {{\left( x\text{ }\text{ }-3 \right)}^{2}}~+\text{ }{{y}^{2}} \right)  \\

\Rightarrow ~{{x}^{2}}~\text{ }-4x\text{ }+\text{ }4\text{ }+\text{ }{{y}^{2}}~=\text{ }4\text{ }\left( {{x}^{2}}~\text{ }-6x\text{ }+\text{ }9\text{ }+\text{ }{{y}^{2}} \right)  \\

\Rightarrow ~3{{x}^{2}}~+\text{ }3{{y}^{2}}~\text{ }-20x\text{ }+\text{ }32\text{ }=\text{ }0  \\

\end{array}\]

Therefore, centre of circle is

\[\left( 10/3,\text{ }0 \right)\] and radius is \[\left( 10/3,\text{ }0 \right)4/9\text{ }or\text{ }2/3\].