Let $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ be end point of diameter of circle, then equation of circle be
$$
\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0
$$

Given, equation of circle is
$$
\begin{aligned}
& x^2+y^2-2 x-6 y+6=0 \\
& \Rightarrow \quad x^2-2 x+1+y^2-6 y+5=0 \\
& \Rightarrow \quad(x-1)^2+(y-1)(y-5)=0 \\
& \Rightarrow(x-1)(x-1)+(y-1)(y-5)=0 \\
&
\end{aligned}
$$

Compare Eqs. (i) and (ii), we obtain end points of diameter be $(1,1)$ and $(1,5)$.
Now, we have to find radius of circle having centre at $(2,1)$ and one of its chord having end point $(1,1)$ and $(1,5)$




From given circle,
$$
\begin{aligned}
x^2+y^2 & =r^2 \\
x=\sqrt{(2-1)^2+(1-3)^2} & =\sqrt{5} \\
y=\sqrt{(1-1)^2+(3-1)^2} & =\sqrt{4} \\
\Rightarrow \quad r^2=x^2+y^2=5+4 & =9 \\
\Rightarrow \quad r & =3
\end{aligned}
$$