How do you write the equation #x^2 + y^2 – x + 2y + 1 = 0# into standard form?

1 Answer
Dec 10, 2015

Answer:

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

Explanation:

Sort the #x#-terms and #y#-terms.

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

Now, complete the square for each variable.

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

Don't forget to balance both sides of the equation. (If you add something to one side, add it to the other side as well.)

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

This is in the standard form of a circle

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

Where:
#(h,k)# is the center and #r# is the length of the radius.

Thus, this is a circle with a center at #(1/2,-1)# and a radius of #1/2#.

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Here's a graph. (Notice the center at #(0.5,-1)# and a diameter of #1#.)