Answer:
Hence, the sphere has a radius of [tex]\sqrt{3}[/tex] and is centered at the point (1,-1,1)
Step-by-step explanation:
We have the equation
[tex]\rho=2cos\theta cos\phi-2sin\theta sin\phi+2cos\phi[/tex]
We have to take into account the relation between coordinates
[tex]\rho=\sqrt{x^2+y^2+z^2}\\x=\rho cos\theta sin\phi\\y=\rho sin\theta sin\phi\\z= \rho cos\phi[/tex]
by substituting we have:
[tex]\rho=2[\frac{x}{\rho}-\frac{y}{\rho}+\frac{z}{\rho}]\\\\\rho^2=2x-2y+2z\\\\x^2+y^2+z^2=2x-2y+2z[/tex]
We have to complete squares:
[tex](x^2-2x+1)+(y^2+2y+1)+(z^2-2z+1)-1-1-1=0\\\\(x-1)^2+(y+1)^2+(z-1)^2=3[/tex]
Hence, the sphere has a radius of [tex]\sqrt{3}[/tex] and is centered at the point (1,-1,1)
hope this helps!!
