How do you determine the center and radius of the following circle and sketch the graph of #x^2+y^2=-4x-3y#?

Begin by writing the equation with the #x# and #y# variables all on one side like this:

#(x^2+4x)+(y^2+3y)=0#

We now apply a process called completing the square to the terms with #x# and again the to the terms having #y#.

Add constants #a# and #b# to the terms in parentheses and also add them to the right side to keep the equation balanced:

#(x^2+4x+a)+(y^2+3y+b)=a+b#

Now find #a# to make the terms in #x# a (completed) square. To do that remember the formula

#(x+u)^2=x^2+2ux+u^2#

To match the #2ux# term with the #4x# that's given in the equation of the circle, put in #u=2#. Then:

#(x+2)^2=x^2+4x+4#

So #a=4# and the completed square is #(x+2)^2#.

Doing the same thing with the terms having #y# the added constant and completed square are found to be:

#b=9/4, (y+3/2)^2=y^2+3y+9/4#

Now put the completed squares into the equation of the circle:

#(x^2+4x+a)+(y^2+3y+b)=a+b#

#(x+2)^2+(y+3/2)=4+9/4={25}/4#

Now compare with the "standard" form for the equation of a circle. If the center is at #(h,k)# and the radius is #r# then:

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

Then #-h=2#, #-k=3/2# so the center is at #(-2,-3/2)#. And the radius is #\sqrt{{25}/4}=5/2#.