Question

How do you convert `x^2+y^2+6x-2y-3=0` to standard form?

Answer 1

`(x-(-3))^2+(y-1)^2 = (sqrt(13))^2`
is the standard form for a circle with center `(-3,1)` and radius `sqrt(13)`

Explanation:

Given `x^2+y^2+6x-2y-3=0`

Re-group the terms as
`x^2+6xcolor(white)("XX")+color(white)("XX")y^2-2ycolor(white)("XX")-3 color(white)("XXX")=0`

Move the constant to the right side and complete the squares
`x^2+6x+9color(white)("XX")+color(white)("XX")y^2-2y+1color(white)("XX")= 3 +9 +1`

Rewrite as squared terms
`(x+3)^2 +(y-1)^2 = 13`

graph{x^2+y^2+6x-2y-3=0 [-8.684, 7.12, -2.95, 4.95]}