JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Circle

First, rewrite the given circle in standard form:
$x^2+y^2-2x-6y+6=0$
$(x-1{)}^2+(y-3{)}^2=1+9-6=4$

So the circle has centre $(1,3)$ and radius $2$.

A diameter of this circle passes through $(1,3)$ and has length $4$. The endpoints of any diameter are at distance $2$ from $(1,3)$.

This diameter is a chord of circle $C$ with centre $(2,1)$. The perpendicular distance from $(2,1)$ to the chord (which passes through $(1,3)$) equals the distance from $(2,1)$ to $(1,3)$:
$d=\sqrt{(2-1{)}^2+(1-3{)}^2}=\sqrt{1+4}=\sqrt{5}$

For a chord of length $4$ (the diameter) in circle $C$ with radius $R$:
$R^2=d^2+(\text{half-chord length}{)}^2=5+4=9$

Therefore $R=3$.