Question 159699
{{{x^2+ y^2 - 2x - 4y - 4 = 0}}}


First, put the constant term on the right:
{{{x^2+ y^2 - 2x - 4y = 4}}}


Divide the coefficient on the first degree x term by 2.  {{{2/2 = 1}}}.


Square the result and add that result to both sides.  {{{1^2=1}}}


{{{x^2+ y^2 - 2x - 4y + 1 = 4 + 1}}}


Repeat the process with the coefficient on the first degree y term


{{{4/2=2}}}, {{{2^2=4}}}


{{{x^2+ y^2 - 2x - 4y + 1 + 4 = 4 + 1 + 4}}}


Rearrange the terms:


{{{x^2-2x+1+y^2-4y+4=9}}}


Now factor the x terms:


{{{x^2-2x+1=(x-1)^2}}}


And factor the y terms:


{{{y^2-4y+4=(y-2)^2}}}


Put the pieces back together:


{{{(x-1)^2+(y-2)^2=9}}}


Now, use the fact that a circle of radius {{{r}}} centered at ({{{h}}},{{{k}}}) has an equation {{{(x-h)^2+(y-k)^2=r^2}}} to determine that your circle is centered at ({{{1}}},{{{2}}}) and has a radius of {{{sqrt(9)=3}}}