Answer:
[tex](x-2)^2+(y+5)^2=36[/tex]
Step-by-step explanation:
Recall that the equation of a circle of radius R centered at [tex](x_0,y_0)[/tex] is given by:
[tex](x-x_0)^2+(y-y_0)^2=R^2[/tex]
For your particular case: R = 6, [tex]x_0[/tex] = 2, and [tex]y_0[/tex] = -5, therefore the equation becomes:
[tex](x-x_0)^2+(y-y_0)^2=R^2\\(x-2)^2+(y-(-5))^2=6^2\\(x-2)^2+(y+5)^2=36[/tex]
