## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$(x-2)^2+(y-1)^2=4$ $r=2$ center is at $(2,1)$
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 $(x-2)^2+(y-1)^2=(2)^2 \implies (x-2)^2+(y-1)^2=4$