this is a circle ... not centered at the origin
complete the squares to find the center and radius
16 [(x^2 - 4x) + (y^2 + 2y)] = -55
16 [(x^2 - 4x + 4) + (y^2 + 2y + 1)] = -55 + [16 * (4 + 1)]
(x - 2)^2 + (y + 1)^2 = 25/16
16x^2 +16y^2-64x+32y+55=0
2 answers
you can tell this is a circle, since x^2 and y^2 both have the same coefficient. So now it's just a matter of rounding up the terms to see where it is and what is its radius.
16x^2 +16y^2-64x+32y+55=0
16x^2-64x + 16y^2+32y = -55
16(x^2-4x) + 16(y^2+2y) = -55
Now complete the squares, making sure the equation still balances.
16(x^2-4x+4) + 16(y^2+2y+1) = -55 + 16*4 + 16*1
16(x-2)^2 + 16(y+1)^2 = 25
(x-2)^2 + (y+1)^2 = 25/16
So now you know that
the center is at (2,-1)
the radius is 5/4
16x^2 +16y^2-64x+32y+55=0
16x^2-64x + 16y^2+32y = -55
16(x^2-4x) + 16(y^2+2y) = -55
Now complete the squares, making sure the equation still balances.
16(x^2-4x+4) + 16(y^2+2y+1) = -55 + 16*4 + 16*1
16(x-2)^2 + 16(y+1)^2 = 25
(x-2)^2 + (y+1)^2 = 25/16
So now you know that
the center is at (2,-1)
the radius is 5/4