Question 161954
{{{x^2 + y^2 = 36}}} Start with the given equation



{{{(x-0)^2 + (y-0)^2 = 36}}} Replace "x" with "x-0" and replace "y" with "y-0".



{{{(x-0)^2 + (y-0)^2 = 6^2}}} Rewrite {{{36}}} as {{{6^2}}}


Since the equation above fits the equation {{{(x-h)^2+(y-k)^2=r^2}}} (which is the equation of any circle) where {{{h=0}}}, {{{k=0}}} and {{{r=6}}}, this shows us that the equation {{{x^2 + y^2 = 36}}} is a circle.




Note: if you aren't entirely sure, solve for "y" and you should get a good idea of the conic section (this isn't foolproof)