Question 754459
the standard formula is:{{{ (x-h)^2+(y-k)^2=r^2}}} where  {{{h}}} and {{{k}}} are the {{{x}}} and {{{y}}} coordinates of the center of the circle, and {{{r}}} is radius 

in your case we have:

{{{(x -2)^2 + (y + 3)^2 = 9}}} 


{{{(x-2)^2 + (y + 3)^2 = 3^2}}} if you compare it to the standard formula, you see that 

{{{h=2}}}, {{{k=-3}}}-the coordinates of the center, 

so, center is at ({{{2}}},{{{-3}}})

{{{r=3}}} and {{{d=6}}}-the radius and diameter 



{{{drawing( 600, 600, -10, 10, -10, 10,circle(2,-3,0.11), graph( 600, 600, -10, 10, -10, 10,-sqrt(-(x-2)^2+9)-3, sqrt(-(x-2)^2+9)-3)) }}}