16x^2 +16y^2-64x+32y+55=0
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
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
The given equation is in the form of a quadratic equation of two variables, x and y. To understand the equation better, let's rearrange it:
16x^2 + 16y^2 - 64x + 32y + 55 = 0
We can reorganize the equation by grouping the x-terms together and the y-terms together:
(16x^2 - 64x) + (16y^2 + 32y) + 55 = 0
Now, we can factor out the common coefficients of x and y:
16(x^2 - 4x) + 16(y^2 + 2y) + 55 = 0
To complete the square for x and y, we need to add and subtract the appropriate constants inside the parentheses:
16(x^2 - 4x + 4 - 4) + 16(y^2 + 2y + 1 - 1) + 55 = 0
Simplifying further:
16((x - 2)^2 - 4) + 16((y + 1)^2 - 1) + 55 = 0
Expanding and rearranging:
16(x - 2)^2 + 16(y + 1)^2 - 64 + 16 + 55 = 0
Combining like terms:
16(x - 2)^2 + 16(y + 1)^2 + 7 = 0
We can see that both (x - 2) and (y + 1) terms are squared. Since squares are always positive or zero, the expression can only be zero if both (x - 2)^2 and (y + 1)^2 are zero. We solve for (x - 2) = 0 and (y + 1) = 0 simultaneously:
x - 2 = 0 ----> x = 2
y + 1 = 0 ----> y = -1
Therefore, the equation 16x^2 + 16y^2 - 64x + 32y + 55 = 0 represents a single point at (2, -1).