SOLUTION: I need help solving this conic in standard form 4x^2+4y^2-4x+2y-1=0

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Question 1044752: I need help solving this conic in standard form
4x^2+4y^2-4x+2y-1=0
Found 3 solutions by Boreal, ikleyn, MathTherapy:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website!
4x^2+4y^2-4x+2y-1=0
4x^2-4x+4y^2+2y=1
4[(x^2+x+y^2+(1/2)y]=1
4(x+(1/2)x+1/4)+(y+(1/4)+(1/16))=1+1+(1/4). Complete the square, adding the constant to both sides.
4(x+(1/2))^2+4(y+(1/4))^2=9/4.
divide both sides by 4
(x+1/2)^2+(y+1/4)^2=9/16
circle with center (-1/2),-1/4) and radius 3/4

Answer by ikleyn(52799)   (Show Source):
You can put this solution on YOUR website!
.
I need help solving this conic in standard form
4x^2+4y^2-4x+2y-1=0
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4x^2 + 4y^2 - 4x + 2y - 1 = 0
4x^2 - 4x + 4y^2 + 2y = 1
4[(x^2 - x + y^2 + (1/2)y] = 1
4(x-(1/2)x+1/4)+(y+(1/4)+(1/16))=1+1+(1/4). Complete the square, adding the constant to both sides.
4(x-(1/2))^2 + 4(y+(1/4))^2 = 9/4.
divide both sides by 4

(x-1/2)^2 + (y+1/4)^2 = 9/16

circle with center (1/2),-1/4) and radius 3/4.

Answer by MathTherapy(10552)   (Show Source):
You can put this solution on YOUR website!
I need help solving this conic in standard form
4x^2+4y^2-4x+2y-1=0

----- Standard, or center-radius form