Answer to Question #245549 in Analytic Geometry for moe

Question #245549

Given the equation of a sphere x2+y2+z24x+2y+2z=5x^{2}+y^{2}+z^{2}-4x+2y+2z=-5

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



  1. r(2,1,1)=1|\mathbf{r}-(2,1,1)|=1
  2. r(2,1,1)=1|\mathbf{r}-(2,-1,-1)|=1
  3. r(2,1,1)=1|\mathbf{r}-(2,-1,1)|=1
  4. r(2,1,1)=1.∣r−(2,−1,1)∣=1.
  5. r(2,1,1)=1.∣r−(2,1,−1)∣=1.
1
Expert's answer
2021-10-21T14:56:33-0400

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)=(2,-1,-1)


radius of sphere =1=1


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


Hence option 2 is correct


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