Question 156163
{{{x^2 + y^2 - 2x - 4y -4 = 0}}} Start with the given equation



{{{x^2- 2x + y^2  - 4y -4 = 0}}} Rearrange the terms (place the "x" and "y" terms together)



{{{x^2- 2x +highlight(1) + y^2  - 4y -4 = highlight(1)}}} Take half of the "x" coefficient and square it to get {{{((-2)/2)^2=(-1)^2=1}}}. Add this to both sides (note: add the one right after the -2x)




{{{x^2- 2x +1 + y^2  - 4y +highlight(4) -4 = 1+highlight(4)}}} Take half of the "y" coefficient and square it to get {{{((-4)/2)^2=(-2)^2=4}}}. Add this to both sides (note: add the four right after the -4y)



{{{x^2- 2x +1 + y^2  - 4y +4 -4 = 5}}} Combine like terms on the right side


{{{(x^2- 2x +1 )+ (y^2  - 4y +4 )-4 = 5}}} Group the terms into two groups of three terms in each



{{{(x-1)^2+ (y^2  - 4y +4 )-4 = 5}}} Factor {{{x^2- 2x +1}}} to get {{{(x-1)^2}}}



{{{(x-1)^2+ (y-2)^2-4 = 5}}} Factor {{{y^2  - 4y +4}}} to get {{{(y-2)^2}}}



{{{(x-1)^2+ (y-2)^2 = 9}}} Add 4 to both sides 



Now we have an equation in the form of {{{(x-h)^2+(y-k)^2=r^2}}} where {{{h=1}}}, {{{k=2}}}, and {{{r=3}}}. Remember, for the general circle {{{(x-h)^2+(y-k)^2=r^2}}}, the center is (h,k) and and the radius is "r"



So for {{{(x-1)^2+ (y-2)^2 = 9}}}, the center is (1,2) and the radius is 3 units