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`.