The circle below is centered at the point (3, 1) and has a radius of length 2.What is its equation?

Answer:[tex]$ x^2 + y^2 - 3x -y +6 = 0 $[/tex]Step-by-step explanation:The equation of a circle with center [tex]$ (h,k) $[/tex] and radius [tex]$ r$[/tex] is given by                  [tex]$ (x - h)^2 + (y - k)^2 = r^2 $[/tex]Here the center is: [tex]$ (3,1) $[/tex] and radius is [tex]$ 2 $[/tex].Therefore, we have:[tex]$ (x - 3)^2 + (y - 1)^2 = 2^2 $[/tex][tex]$ \implies x^2 -3x +9 +y^2 -y +1 = 4 $[/tex][tex]$ \implies x^2 + y^2 -3x -y +6 = 0 $[/tex]