Question 581656
Note: the origin is the point (0,0)


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



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



{{{(3-0)^2+(5-0)^2=r^2}}} Plug in {{{x=3}}} and {{{y=5}}} (this is the point that lies on the circle, which is in the form (x,y) ).



{{{(3)^2+(5)^2=r^2}}} Combine like terms.



{{{9+25=r^2}}} Square each term.



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



So because  {{{h=0}}}, {{{k=0}}}, and {{{r^2=34}}}, this means that the equation of the circle with center (0,0) that goes through the point (3,5) is 



{{{(x-0)^2+(y-0)^2=34}}}.



which simplifies to {{{x^2+y^2=34}}}.



So the answer is {{{x^2+y^2=34}}}