EXPLANATION
We already know that the equation of the circle centered at (h,k) and that contains the point (x,y) is as follows:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
Where h=2 and y=1.
Replacing terms:
[tex](x-2)^2+(y-1)^2=r^2[/tex]
Since (3,-2) is on the graph, we have:
[tex](3-2)^2+(1-1)^2=r^2[/tex]
Subtracting numbers:
[tex]1^2+0=r^2\text{ }\longrightarrow\text{ 1=r\textasciicircum{}2}\longrightarrow1=r\text{ }[/tex]
As r=1 our equation is as follows:
[tex](x-2)^2+(y-1)^2=1[/tex]
