If two circles touching both the axes are passing through (2, 3) then length of their common chord is 
 

$$\begin{gathered} \mathrm{From} \mathrm{the} \mathrm{figure}, \\ \mathrm{tan\theta }=\frac{{\displaystyle \frac{3}{2}}-1}{1+\frac{3}{2}} =\frac{1}{5} \\ \mathrm{In} \mathrm{\Delta ADO},\mathrm{sin\theta }=\frac{\mathrm{AD}}{\mathrm{OA}} \\ \mathrm{We} \mathrm{have} \mathrm{tan\theta }=\frac{1}{5} \\ \Rightarrow \frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}=\frac{1}{5} \\ 5\mathrm{sin\theta }=\mathrm{cos\theta } \\ \mathrm{using} \mathrm{identity} {\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}=1 \\ \mathrm{we} \mathrm{get} {\sin }^2\mathrm{\theta }+25{\sin }^2\mathrm{\theta }=1 \Rightarrow 26{\sin }^2\mathrm{\theta }=1 \\ \therefore \mathrm{sin\theta }= \frac{1}{\sqrt{26}} \\ \Rightarrow \frac{1}{\sqrt{26}}=\frac{\mathrm{AD}}{\mathrm{OA}} \Rightarrow \frac{1}{\sqrt{26}}=\frac{\mathrm{AD}}{13} \\ \Rightarrow \mathrm{AD}=\frac{13}{\sqrt{26}} \\ \Rightarrow \mathrm{AD}=\frac{1}{\sqrt{2}} \\ \mathrm{Thus} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{common} \mathrm{chord} \mathrm{is} 2\mathrm{AD}=2\times \frac{1}{\sqrt{2}}=\sqrt{2} \end{gathered}$$