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write the standard form of the equation of the circle: x^2+y^2-2x+6y+6=0
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the form u wrote is the standard form:)
no. standard form is: (x-h)^2+(y-k)^2=r^2
Yeah you have to complete the square, you got your notes on that nomss?
might have to complete the square
there is an example in the book
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(x-1)^2+(y+3)^2=2^2
(x^2-2x+1) +(y^2+6y+9)=-6+9+1
I have this: (x^2-2x+1) +(y^2+6y+9)=-6+9+1, what now? I don't think I can do (x+1)^2 + (y+3)2 = 4?
right (x-1)^2+(y+3)^2=2^2
cool, i see, thanks.
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