Question 1044752
.
I need help solving this conic in standard form

4x^2+4y^2-4x+2y-1=0
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


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.