Question 1106895
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The standard form equation of the circle with radius r and center (h,k) is

{{{(x-h)^2}}} + {{{(y-k)^2}}} = {{{r^2}}}.         (1)     ( <U>MEMORIZE it !</U> )


    In your case it is 

    {{{(x-0)^2}}} + {{{(y-3)^2}}} = {{{3^2}}},     (2)    or

    {{{x^2}}} + {{{(y-3)^2}}} = 9.


To get the general form equation, simply open the parentheses in equation (1).


    In your case it will give you

    {{{x^2 + y^2 - 6y + 9}}} = 9,     or, equivalently,

    {{{x^2 + y^2 - 6y}}} = 0.


<U>Answer</U>.  Standard form equation is  {{{x^2}}} + {{{(y-3)^2}}} = 9.


         General form equation is  {{{x^2 + y^2 - 6y}}} = 0.
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