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

Equation of the circle C is ${\left(x-2\right)}^2+{\left(y-1\right)}^2=r^2$

$\Rightarrow x^2+y^2-4x-2y+5-r^2=0$

Equation of the common chord is $\left(x^2+y^2-4x-2y+5-r^2\right)-\left(x^2+y^2-2x-6y+6\right)=0$

$\Rightarrow 2x-4y+r^2+1=0$

If it is a diameter of the second circle, it passes through the centre $\left(1,3\right)$  of the circle

So $2-4\times 3+r^2+1=0$ $\Rightarrow r^2=9$ $\Rightarrow r=3$