Answer
$(x-2)^2+(y-1)^2=4$
$r=2$
center is at $(2,1)$
Work Step by Step
The standard form of a equation of a circle is:
$(x-h)^2+(y-k)^2=r^2$ ...(1)
where $(h,k)=$ center and $r$= radius.
The center of a circle is at the point $(2,1)$ and this means that, $h=2$ and $k=1$,
To calculate the radius $r$ use distance $d$ formula between two points$(x_1,y_1)$ and $(x_2,y_2)$, we have:
$r=d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
Plug the $x$ and $y$-coordinates of the given points to obtain:
$d=\sqrt{(2-0)^2+(1-1)^2}=\sqrt {4+0}=2$
With $r=2$, $h=2$, and $k=1$, the equation of the given circle is
$(x-2)^2+(y-1)^2=(2)^2 \implies (x-2)^2+(y-1)^2=4$
