How would i show that the equation is a circle? And how would i find it's center and radius?

Question

4x^2+ 4y^2 + 4z^2 - 8x + 16y=1

sorry, i need to correct the very important key word. it's sphere. not circle -___-

Answer 1

i don't know that this would be a circle, you have a z^2 term which would make this 3 space not 2 space. Are you sure the 4z^2 is supposed to be there? If not, then you would have

4x^2 + 4y^2 -8x +16y = 1

What you should do is get the x's together and the y's together, then you can complete the square on both. But first double check and make sure the equation is correct.

Answer 2

Since the coefficients of x^2, y^2, and z^2 are the same, and 1 is positive, you have a sphere:
4x^2 + 4y^2 + 4z^2 - 8x + 16y = 1
x^2 - 2x + y^2 + 4y + z^2 = 1/4
x^2 - 2x + 1 + y^2 + 4y + 4 + z^2 = 1/4 + 1 + 4
(x - 1)^2 + (y + 2)^2 + z^2 = (1 + 4 + 16)/4
(x - 1)^2 + (y + 2)^2 + z^2 = 21/4
center: 1, -2, 0
radius: (1/2)√21 ≈ 2.291288

Answer 3

4x² + 4y² + 4z² - 8x + 16y = 1
Divide by 4
x² + y² + z² - 2x + 4y = 1/4
Complete the squares:
x² - 2x + 1 + y² + 4y + 4 + z² = 21/4
(x-1)² + (y+2)² + z² = 21/4

Center: (1, -2, 0)
Radius: sqrt(21/4) = sqrt(21)/2