Answer
$\color{blue}{(x-1)^2+(y-2)^2=4}$
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 $(1, 2)$. Substituting $1$ to $h$ and $2$ to $k$ gives that tentative equation of the circle:
$(x-1)^2+(y-2)^2=r^2$
To find the value of $r$, substitute the $x$ and $y$ coordinates of the point on the circle $(1, 0)$ to obtain:
$(x-1)^2+(y-2)^2=r^2
\\(1-1)^2+(0-2)^2=r^2
\\0^2+(-2)^2=r^2
\\0+4=r^2
\\4=r^2
\\\pm\sqrt{4} = \sqrt{r^2}
\\\pm\sqrt2 = r$
Since a radius cannot be negative, then $r=2$.
Therefore, the standard form of the circle's equation is:
$(x-1)^2+(y-2)^2=2^2
\\\color{blue}{(x-1)^2+(y-2)^2=4}$
