How do you write # x^2 – x + y^2 + y = 0# in standard form and why type of conic is it?

1 Answer
Dec 3, 2015

Answer:

Circle: #(x-1/2)^2+(y+1/2)^2=1/2#

Explanation:

Complete the square for both the #x# and #y# parts.

#color(blue)((x^2-x+?))+color(red)((y^2+y+?))=0#

#color(blue)((x^2-x+1/4))+color(red)((y^2+y+1/4))=1/4+1/4#

We add #1/4# to both the #x# and #y# portions because they will factor into #(x-1/2)^2# and #(y+1/2)^2#, respectively. Don't forget to add to the right side the exact same numbers you added to the left so the equation remains balanced.

#(x-1/2)^2+(y+1/2)^2=1/2#

This is a circle, since the standard form of a circle is:

#(x-h)^2+(y-k)^2=r^2#

There's a center at #(1/2,-1/2)# and a radius of #sqrt2/2#.