How do you find the standard form of $x^{2} + y^{2} + 2 x - 6 y - 6 = 0$ $x^2 + y^2 + 2x - 6y - 6 = 0$ 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)

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

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

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

How do you find the standard form of x^2 + y^2 + 2x - 6y - 6 = 0x2+y2+2x−6y−6=0x^2 + y^2 + 2x - 6y - 6 = 0 and what kind of a conic is it?