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

1 Answer
Dec 5, 2015

Answer:

#(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)#