How do you graph # (x+3)^2 + (y-2)^2 = 25#?

The function is a circular function:

The general form of a circular function can be expressed as:

#(x-h)^2+(y-k)^2=r^2#

Where the centre point of the graph is present at the point #(h,k)#
and the solution (the number after the = sign) is the radius of the circle squared.

Therefore, from your function:

#(x+3)^2+(y-2)^2=25#

We can determine that:

The centre point of the function is present at the point #(-3,2)# and the radius is #sqrt25=5#

For any function, #x# incepts where #y# = 0

Therefore, by substituting #y=0# we get:

#(x+3)^2+(-2)^2=25#

By simplifying this and solving for #x# we get:

#(x+3)^2+4=25#
#(x+3)^2=25-4#
#(x+3)^2=21#
#(x+3)=+-sqrt21#
Remember that any real square has two solutions (a positive and negative), hence the #+-sqrt21#
#x=+-sqrt21-3#
Therefore, two #x# intercepts are present:
One at: #x=-sqrt21-3#
The other at: #x=+sqrt21-3#

For any function, #y# intercepts where #x# = 0

Therefore, if we substitute #x# = 0 into your equation, we get:

#(0+3)^2+(y-2)^2=25#

By simplifying and solving for #y# we get:

#9+(y-2)^2=25#
#(y-2)^2=16#
Again, remember that any real square has two solutions (a positive and negative), hence the #+-sqrt16#
#(y-2)=+-sqrt16#
#(y-2)=+-4#
#y=+-4+2#
Therefore, two #y# intercepts are present:
One at: #y=-2# calculated as: #[-4+2]#
The other at: #y=6# calculated as: #[+4+2]#

To summarise:

If we plot all of our points on the graph, we get:

Centre Point at: #(-3,2)#

Radius of: 5

#x# Intercepts at: #x= +sqrt21-3# and #x=-sqrt21-3#

#y# Intercepts at: #y=-2# and #y=6#
graph{(x+3)^2+(y-2)^2=25 [-21.42, 18.58, -7.32, 12.68]}