Question

What is the standard form of `x^2 - 2x + y^2 + 4y = 11`?

Answer 1

`(x-1)^2+(y-(-2))^2=(sqrt(6))^2`

Explanation:

Recognizing the given equation `x^2-2x+y^2+4y=11`
as the equation of a circle
and
knowing that the standard form of a circle with center `(a,b)` and radius `r` is
`color(white)("XXX")(x-a)^2+(y-b)^2 = r^2`

We need to complete the squares for each of `x` and `y` in the given form:
`color(white)("XXX")(x^2-2x+1)-1 +(y^2+4y+4)-4 = 11`

`color(white)("XXX")(x-1)^2 + (y+2)^2 -5 = 11`

`color(white)("XXX")(x-1)^2+(y+2)^2 = 6`

Normally I would consider the above "close enough" to standard form;
however to be completely accurate:
`color(white)("XXX")(x-(1))^2+(y-(-2))^2= (sqrt(6))^2`
with center at `(1,-2)` and radius `sqrt(6)`