Question 530466
{{{(x-h)^2+(y-k)^2=r^2}}} Start with the general equation of a circle.



{{{(x-0)^2+(y-4)^2=r^2}}} Plug in {{{h=0}}} and {{{k=4}}} (since the center is the point (h,k) ).



{{{(0-0)^2+(0-4)^2=r^2}}} Plug in {{{x=0}}} and {{{y=0}}} (this is the point that lies on the circle, which is in the form (x,y) ).



{{{(0)^2+(-4)^2=r^2}}} Combine like terms.



{{{0+16=r^2}}} Square each term.



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



So because  {{{h=0}}}, {{{k=4}}}, and {{{r^2=16}}}, this means that the equation of the circle with center (0,4) that goes through the point (0,0) is 



{{{(x-0)^2+(y-4)^2=16}}}.



which simplifies to 



{{{x^2+(y-4)^2=16}}}



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