We are given equation of sphere x^2 + y^2 + z^2 + 6x − 2y − 6z = 17.
We know , standard form of a sphere (x−a)^2+(y−b)^2+(x−c)^2=r^2.
Where centered at (a,b,c) ( a , b , c ) with radius r.
Let us convert given equation of sphere in standard form by completing the square.
Writing x, y and z terms seprately.
(x^2+6x) + (y^2-2y) + (z^2-6z) = 17.
We have coefficents of x, y and z are 6, -2 and -6.
First we will find half of those coefficents and then square them.
We get 6/2 =3, -2/2 =-1 and -6/2=-3.
Now, squaring we get
(3)^2 = 9, (-1)^2 = 1 and (-3)^2 = 9.
Adding those numbers inside first, second and third parentheses and adding same numbers on right side.
(x^2+6x+9) + (y^2-2y+1) + (z^2-6z+9) = 17+9+1+9.
Now, finding perfect square of each, we get
(x+3)^2 + (y-1)^2 + (z-3)^2 = 36.
We could write 36 as 6^2.
So, final answer would be
(x+3)^2 + (y-1)^2 + (z-3)^2 = 6^2 is the standard form of a sphere equation.
