## College Algebra (10th Edition)

center: $(2, 1)$ radius = $2$ standard form of the equation:$\color{blue}{(x-2)^2+(y-1)^2=4}$
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: $(x-2)^2+(y-1)^2=2^2 \\\color{blue}{(x-2)^2+(y-1)^2=4}$