Answer
center: $(2, 1)$
radius = $2$
standard form of the equation:$\color{blue}{(x-2)^2+(y-1)^2=4}$
Work Step by Step
RECALL:
The standard form of a circle's equation is:
$(x-h)^2 +(y-k)^2=r^2$
where $r$ = radius and $(h, k)$ is the center.
The given circle has its center at $(2, 1)$.
Substituting $2$ to $h$ and $1$ to $k$ gives the tentative equation of the circle:
$(x-2)^2+(y-1)^2=r^2$
To find the value of $r$, substitute the $x$ and $y$ coordinates of the point on the circle $(0, 1)$ to obtain:
$(x-2)^2+(y-1)^2=r^2
\\(0-2)^2+(1-1)^2=r^2
\\(-2)^2+0^2=r^2
\\4+0=r^2
\\4=r^2
\\\pm\sqrt{4} = \sqrt{r^2}
\\\pm2 = r$
Since a radius cannot be negative, $r=2$.
Therefore, the standard form of the circle's equation is:
$(x-2)^2+(y-1)^2=2^2
\\\color{blue}{(x-2)^2+(y-1)^2=4}$