How do you find the standard form of $x^2 + y^2 + 2x - 6y - 6 = 0$ and what kind of a conic is it?

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Answer 1 · 2018-06-01T17:10:29

Circle, radius 4 and centre (-1,3)

$x^2+2x+y^2-6y=6$

$(x+1)^2-1+(y-3)^2-9=6$

$(x+1)^2+(y-3)^2=16$

How do you find the standard form of x^2 + y^2 + 2x - 6y - 6 = 0x^2 + y^2 + 2x - 6y - 6 = 0 and what kind of a conic is it?