Question

x^2+y^2+z^2+3x-4z+1=0
(x^2+3x)+y^2+(z^2-4z)=-1
where do I go from here to get the radius and center for the sphere

(x-x0)^2+(y-y0)^2+(z-z0)^2=a^2

Answer 1

from \(x^2+y^2+z^2+3x-4z+1=0\)
to \((x^2+3x)+y^2+(z^2-4z)=-1\)

you need to complete the squares for x,y,z

so, for x:

\( x^2+3x= (x + \frac{3}{2})^2 - \frac{9}{4} \)

then add them all up :p

Answer 2

where did 3/2 come from.

Answer 3

what is
\( (x + \frac{3}{2})^2\)??

Answer 4

oh okay thanks