Answer
Graph of $
(x-2)^2+(y-3)^2=4
$
Work Step by Step
Using $(x-h)^2+(y-k)^2=r^2$ or the Center-Radius Form, the given equation, $
(x-2)^2+(y-3)^2=4
,$ is equivalent to
\begin{align*}\require{cancel}
(x-2)^2+(y-3)^2=2^2
.\end{align*}
Since the center of a circle is given by $(h,k),$ then the center of the circle with the equation above is $
(2,3)
$.
Since the radius is given by $r$, then the radius of the circle above is $
2
$. Moving $r$ units to the right of the center results to the point $
(4,3)
$. Moving $r$ units above the center results to the point $
(2,5)
$. Moving $r$ units to the left of the center results to the point $
(0,3)
$. Moving $r$ units below the center results to the point $
(2,1)
$. Connecting these points give the graph of the circle defined by the equation $
(x-2)^2+(y-3)^2=4
$.
