Question 530469
{{{(x-h)^2+(y-k)^2=r^2}}} Start with the general equation of a circle.



{{{(x-4)^2+(y-6)^2=r^2}}} Plug in {{{h=4}}} and {{{k=6}}} (since the center is the point (h,k) ).



{{{(2-4)^2+(2-6)^2=r^2}}} Plug in {{{x=2}}} and {{{y=2}}} (this is the point that lies on the circle, which is in the form (x,y) ).



{{{(-2)^2+(-4)^2=r^2}}} Combine like terms.



{{{4+16=r^2}}} Square each term.



{{{20=r^2}}} Add.



So because  {{{h=4}}}, {{{k=6}}}, and {{{r^2=20}}}, this means that the equation of the circle with center (4,6) that goes through the point (2,2) is 



{{{(x-4)^2+(y-6)^2=20}}}.