Question

Write the equation of the sphere in standard form. 2x2 + 2y2 + 2z2 = 12x − 24z + 1

Answer 1

The standard equation of a sphere with center C(a, b, c) and radius r is:

$(x-a)^2 + (x-b)^2 + (x-c)^2 = r^2$
Now, the given equation is :

$2x^2 + 2y^2 + 2z^2 = 12x - 24z + 1 \\ 2x^2 + 2y^2 + 2z^2 - 12x + 24z - 1 = 0$

combine the x, y and z terms and complete the square for each variable 
$(2x^2 - 12x) + 2y^2 + (2z^2 + 16z) = 1 \\ (x^2 - 6x) + y^2 + (z^2 + 8z) = 1/2$


complete square for x by adding 9 to both sides 

we do not complete square for y because there is not a translation on that axis 

complete square for z by add 16 to both sides 
$(x^2 - 6x + 9) + y^2 + (z^2 - 8x + 16) = 1/2 + 9 + 16 \\ (x - 3)^2 + y^2 + (z - 4)^2 = 25 1/2$

Answer 2

The standard equation of a sphere with center C(a, b, c) and radius r is:

$(x-a)^{2}+ (y-b)^{2}+ (z-c)^{2}=r^{2}$


We have the equation: 

$2x^2+2y^2+2z^2=12x-24z+1$

which can be regrouped to fit the standard form, using completing the square as follows:

first divide both sides by 2

$x^2+y^2+z^2=6x-12z+ \frac{1}{2} \\\\(x^2-6x)+y^2+(z^2+12z)= \frac{1}{2}\\\\(x^2-2\cdot3\cdot x)+y^2+(z^2+2 \cdot6 \cdot z)= \frac{1}{2}\\\\(x^2-2\cdot3\cdot x+3^2)-3^2+y^2+(z^2+2 \cdot6 \cdot z+6^2)-6^2= \frac{1}{2}\\\\(x-3)^2+y^2+(x+6)^2=9+36+0.5\\\\(x-3)^2+y^2+(x+6)^2=45.5\\\\\(x-3)^2+y^2+(x+6)^2=( \sqrt{45.5} )^2$


Answer: $(x-3)^2+y^2+(x+6)^2=( \sqrt{45.5} )^2$