Write the equation of the sphere in standard form. X2 + y2 + z2 + 6x - 2y - 6z = 17

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.