## Ashley1nOnly one year ago 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

1. IrishBoy123

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

2. Ashley1nOnly

where did 3/2 come from.

3. IrishBoy123

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

4. Ashley1nOnly

oh okay thanks