[tex]Equation\ of\ circle\\\\\
r^2=(x-a)^2+(y-b)^2\\\\
a,b- \ coordinates\ of\ center\ of\ circle\\\\\
(a,b)=(3,5)\\\\
r^2=(x-3)^2+(y-5)^2\\\\\
Substituting\ point\ (0,0)-origin\\\\
r^2=(0-3)^2+(0-5)^2\\\\
r^2=(-3)^2+(-5)^2\\\\
r^2=9+25=36\\
r=6\\\\
Answer:\ \ 36=(x-3)^2+(y-5)^2[/tex]
