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Mathematics 11 Online
OpenStudy (anonymous):

write the standard form of the equation of the circle: x^2+y^2-2x+6y+6=0

OpenStudy (anonymous):

the form u wrote is the standard form:)

OpenStudy (anonymous):

no. standard form is: (x-h)^2+(y-k)^2=r^2

OpenStudy (anonymous):

Yeah you have to complete the square, you got your notes on that nomss?

OpenStudy (anonymous):

might have to complete the square

OpenStudy (anonymous):

there is an example in the book

OpenStudy (anonymous):

(x-1)^2+(y+3)^2=2^2

OpenStudy (anonymous):

(x^2-2x+1) +(y^2+6y+9)=-6+9+1

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

right (x-1)^2+(y+3)^2=2^2

OpenStudy (anonymous):

cool, i see, thanks.

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