The radius of circle having centre at (2, 1) and whose one of the chord is diameter of the circle $x^2+y^2-2x-6y+6=0$ 
 

$$\begin{gathered} Given circle x^2+y^2-2x-6y+6=0 \\ radius=\sqrt{1+9-6}=2 \\ \therefore diameter=2\times 2=4 \\ Length of chord =y \\ The midpoint of chord =(1,3) (\because midpoint is the centre of the other circle) \\ k= Dis\tan ce between (2,1) and (1,3) \\ =\sqrt{{(2-1)}^2+{(1-3)}^2}=\sqrt{5 } \\ From diagram \\ \Rightarrow k^2+2^2=r^2 \\ \Rightarrow 5+4=r^2 \\ \Rightarrow r=3 \end{gathered}$$