Question

Find the centre and the radius of the sphere
$x^2+y^2+z^2+2x-4y-6z+5=0$, & $2x^2+2y^2+2z^2-2x+4y+2z+3=0$.

• A. $C_1=(1,2,3),r_1=3,C_2=(\frac{1}{2},-1,-\frac{1}{2}),r_2=3$
• B. $C_1=(1,2,3),r_1=3,C_2=(\frac{1}{2},2,\frac{1}{2}),r_2=\sqrt{3}$
• C. $C_1=(1,-2,-3),r_1=3,C_2=(-\frac{1}{2},2,\frac{1}{2}),r_2=3$
• D. $C_1=(1,2,3),r_1=\sqrt{3},C_2=(-\frac{1}{2},2,\frac{1}{2}),r_2=3$
  

Answer 1

Answer: (A)

$C_1=(1,2,3),r_1=3,C_2=(\frac{1}{2},-1,-\frac{1}{2}),r_2=3$

Sphere : $x^2+y^2+z^2-2x-4y-6z+5=0$

$\Rightarrow (x^2-2x+1)+(y^2-4y+4)+(z^2-6z+9)=9$

$\Rightarrow (x-1{)}^2+(y-2{)}^2+(z-3{)}^2=3^2$

$\therefore$ Centre = $(1,2,3)$ radius = $3$

Sphere : $2x^2+2y^2+2z^2-2x+4y+2z+3=0$

$\Rightarrow x^2+y^2+z^2-x+2y+z+\frac{3}{2}=0$

$\Rightarrow (x^2-x+\frac{1}{4})+(y^2+2y+1)+(z^2+z+\frac{1}{4})=9$

$\Rightarrow (x-\frac{1}{2}{)}^2+(y+1{)}^2+(z+\frac{1}{2}{)}^2=9$

$\therefore$ Centre = $(\frac{1}{2},-1,\frac{-1}{2})$ radius = $3$