Question

How do you write `2x² + 2y² - 6x + 4y + 1 = 0` in standard form?

Answer 1

Standard circle form: `(x-3/2)^2+(y+1)^2=(sqrt(11)/2)^2`

Explanation:

Note that the given equation is in standard polynomial form
so I have assumed you need it converted to standard circle form.

`2x^2+2y^2-6x+4y+1=0`

`rarr (2x^2-6xcolor(white)("XXXX"))+(2y^2+4ycolor(white)("XXXX")) = -1`

`rarr 2(x^2-3xcolor(red)(+(3/2)^2))+2(y^2+4ycolor(blue)(+1))=-1color(red)(+2(3/2)^2)color(blue)(+2(1))`

`rarr 2(x-3/2)^2+2(y+1)^2=11/2`

`rarr (x-3/2)^2+(y+1)^2 = 11/4`

`rarr (x-3/2)^2+(y+1)^2=(sqrt(11)/2)^2`
which is standard circle form for a circle with center `(3/2,-1)` and radius `sqrt(11)/2`