Question 970939
Complete the Squares, and make use of the standard form of the finished equation.


{{{x^2-3x+y^2-y+4=0}}}-----THE MISTAKE HERE-.
{{{x^2-3x+(3/2)^2+y^2-y+(1/2)^2+4-(3/2)^2-(1/2)^2=0}}}
{{{(x-3/2)^2+(y-1/2)^2+4-9/4-1/4=0}}}
{{{(x-3/2)^2+(y-1/2)^2=10/4-4}}}
{{{(x-3/2)^2+(y-1/2)^2=-3/2}}}------This will not work.  This is not a real circle.
{{{r^2=-3/2}}}---No good.


Look for any mistakes in your given equation or in my steps.  I found none in mine but examine if you want.
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EDIT:  FOUND THE MISTAKE IN MY WORK.