Question 648986
the general formula of the circle is {{{(x-h)^2+(y-k)^2=r^2}}} where {{{ h}}} and {{{k}}} are the {{{x}}} and {{{y}}} coordinates of the {{{center}}} of the circle , and {{{r}}} is the {{{radius}}}

you are given {{{(x-3)^2+(y-1)^2=36}}}

if you compare your equation to the general formula of the circle, you can see that {{{h=3}}} and {{{k=1}}};so, the {{{center}}} is at ({{{3}}},{{{1}}})

and {{{r^2=36}}}...->... {{{r=6}}}


*[invoke Plot_a_circle 3, 1, 6]