How do you graph #(x+3)^2+(y-2)^2=9#?

1 Answer
Jun 17, 2017

Please see below.

Explanation:

The equation #(x+3)^2+(y-2)^2=9# can be written as

#sqrt((x+3)^2+(y-2)^2)=3#

which indicates that the point #(x,y)# moves so that it is always at a distance of #3# from a point #(-3,2)#

Hence graph of #(x+3)^2+(y-2)^2=9# can be drawn by drawing a circle whose centre is #(-3,2)# and radius is #3#. It appears as shown below.
graph{((x+3)^2+(y-2)^2-9)((x+3)^2+(y-2)^2-0.03)=0 [-12.83, 7.17, -2.76, 7.24]}