circles

find the center and radius:
x^2+y^2-6x+4y-12=0
x^2-6x+__+y^2+4y+__=12
x^2-6x+9+y^2+4y+4=12+9+4
(x-3)^2+(y+2)^2=25
C: (3,-2)  R=5

To start off this problem you have it in standard form.  Standard form for a circle is (k-h)^2+(y-k)^2=r^2. Then, for this problem they are asking you for the center and radius.  To find the radius you can half the diameter found by using 2 points on the outside through the center.  The equation we have is in standard form, so now you have to half the middle term, then square it and add that number to both sides.  So we took 6 and divided it by 2, then squared that number and added it to both sides of the equation.  You do the same thing with the 4 and add it to both sides.  Now your equation looks like (x-3)^2+(y+2)^2=25.  To find the center you take the opposite signs of the numbers and parentheses and make it into a point.  So your center is (3,-2).  To find the radius you just take the 25 and square root it to get 5.