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

1 Answer
Dec 19, 2015

Answer:

#(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]}