Question 975231
Given the equation of the circle is {{{x^2-4x+y^2-6y=3}}}.
x^2-4x + 4 + y^2-6y+9 = 3 +4+9
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(x-2)^ + (y-3)^2 = 16
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Find the equation of the circle in the form{{{(x-a)^2+(y-b)^2=r^2}}}. 
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Hence find out the 
centre:: (2,3)
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area = pi*4^2 = 16pi square units
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circumference of the circle = 2*pi*4 = 8pi units 
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Cheers,
Stan H.
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