What is the radius of the circle $B$? If one of the diameters of the circle $A$ with equation $x^2+y^2-2x-6y+6=0$ is a chord to the circle $B$ with centre $(2, 1)$ then the radius of the circle $B$ is
$(1)\ \sqrt 3$
$(2)\ \sqrt 2$
$(3)\ 3$
$(4)\ 2$
If I draw a perpendicular from $(2,1)$ to that diameter of $A$ then since it is the chord of $B$ the foot of the perpendicular should be $(1,3)$ since the centre of $A$ is the midpoint of that diameter.So the perpendicular distance of that chord of $B$ (equivalently the diameter of $A$) from the centre of $B$ is $\sqrt 5$ and hence the radius of $B$ is  $\sqrt {2^2 + (\sqrt 5)^2} =\sqrt 9 = 3$. So $(3)$ is the correct answer.
Is it correct at all? Please verify it.
Thank you in advance.

Yes, it is correct. Note that the equation $x^2+y^2-2x-6y+6=0$ can be rewritten as
$$(x-1)^2+(y-3)^2=1+9-6=2^2$$
so the circle $A$ has centre $c_A=(1,3)$ and radius $r_A=2$. Therefore the radius of $B$ is given by
$$r_B=\sqrt{r_A^2+d(c_A,c_B)^2}=\sqrt {2^2 + (1-2)^2+(3-1)^2} =\sqrt 9 = 3.$$
