Answer:
This a circle centered at the point [tex](-6,-5)[/tex], and of radius "3" as it is shown in the attached image.
Step-by-step explanation:
Recall that the standard formula for a circle of radius "R", and centered at the point [tex](x_0,y_0)[/tex] is given by:
[tex](x-x_0)^2+(y-y_0)^2=R^2[/tex]
Therefore, in our case, by looking at the standard equation they give us, we extract the following info:
1) [tex]R^2 = 9\,\,\,then\,\,\,R=3[/tex] since the radius must be a positive number and ([tex]R=-3[/tex]) is not a viable answer.
2) [tex]x_0=-6[/tex] for ( [tex]x-x_0[/tex]) to equal [tex](x+6)[/tex]
3) [tex]y_0=-5[/tex] for ( [tex]y-y_0[/tex]) to equal [tex](y+5)[/tex]
Therefore, we are in the presence of a circle centered at the point [tex](-6,-5)[/tex], and of radius "3". That is what we draw as seen in the attached image.
