If common chord of the circle $C$ with centre at $(2, 1)$ and radius $r$ and the circle $x^2+y^2–2x–6y+6=0$ is a diameter of the second circle, then the value of $r$ is

Equation of $C$ is $(x-2{)}^2+(y-1{)}^2=r^2$
or $x^2+y^2-4x-2y=r^2-5$
Equation of the common chord is
$\begin{array}{l}{x}^2+y^2-4x-2y-\left(r^2-5\right)-\left[x^2+y^2-2x-6y+6\right]=0\\ \Rightarrow 2x-4y+r^2+1=0\end{array}$
Since it is a diameters of the second circle, the centre $(1, 3)$ lies on it.

$\Rightarrow 2-12+r^2+1=0\Rightarrow r=3.$