Answer to Question #245549 in Analytic Geometry for moe

Question

Given the equation of a sphere $x^{2}+y^{2}+z^{2}-4x+2y+2z=-5$

Which of the following is the vector equation of the sphere above?

1. $|\mathbf{r}-(2,1,1)|=1$
2. $|\mathbf{r}-(2,-1,-1)|=1$
3. $|\mathbf{r}-(2,-1,1)|=1$
4. $∣r−(2,−1,1)∣=1.$
5. $∣r−(2,1,−1)∣=1.$

Answer 1

Given the equation of the sphere

x^2+y^2+z^2-yx+2y+2z=-5\\\implies xx^2-4x+y^2+2y+z^2+2z=-5\\\implies (x^2-4x+2^2-2^2)+(y^2+2y+1-1)+(z^2+2Z+1-1)=-5\\ [(x-2)^2-4]+[(y+1)^2-1]+[(z+1)^2-1]=-5\\\implies (x-2)^2+(y+1)^2+(z+1)^2-6=-5\\\implies (x-2)^2+(y+1)^2+(z+1)^2=1\\

centre of sphere $=(2,-1,-1)$

radius of sphere $=1$

vector form is $|r-(2,-1,-1)|=1$

Hence option 2 is correct