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

1 Answer
Oct 26, 2016

Answer:

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#