Question

Write the equation of the sphere in standard form. x2 + y2 + z2 + 12x − 4y + 2z + 37 = 0 (x+6)2+(y−2)2+(z+1)2=4 find its center and radius.

Answer 1

The given equation of sphere is

$x^2 + y^2 + z^2 + 12x - 4y + 2z + 37 = 0$

We have to use completing the square method , and in that method, we have to divide 12x ,-4y,2z by 2, square the results, and add the reult obtained after squaring to both sides. That is

$x^2 + y^2 + z^2 + 12x - 4y + 2z+36+4+1  = -37+36+4+1$

$x^2 +12x+36+ y^2-4y+4 + z^2 +2z+1 = 4$

$(x+6)^2 +(y-2)^2+(z+1)^2 = 4$

So here the center is (-6,2,-1) and radius is 2 .