Question

Find the equation of the sphere through the points $(4,-1,2),(0,-2,3),(1,5,-1)$ and $(2,0,1)$.

• A. $x^2+y^2+z^2+4x+14y+22z-25=0$
• B. $x^2+y^2+z^2-4x-14y-22z+25=0$
• C. $x^2+y^2+z^2-2x-17y-11z+25=0$
• D. $x^2+y^2+z^2+2x+17y+11z-25=0$
  

Answer 1

Answer: (B)

$x^2+y^2+z^2-4x-14y-22z+25=0$

Equation of sphere :

$(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=r^2$

where $(a,b,c)$ is the centre

r is the radius

For point $(4,-1,2)$

$\Rightarrow (4-a{)}^2+(-1-b{)}^2+(2-c{)}^2=r^2-(1)$

For point $(0,-2,3)$

$\Rightarrow (0-a{)}^2+(-2-b{)}^2+(3-c{)}^2=r^2-(2)$

For point $(1,-5,-1)$

$\Rightarrow (1-a{)}^2+(-5-b{)}^2+(-1-c{)}^2=r^2-(3)$

For point $(2,0,1)$

$\Rightarrow (2-a{)}^2+(0-b{)}^2+(1-c{)}^2=r^2-(4)$

On solving 4 equation,

we get

$x^2+y^2+z^2-4x-14y-22z+25=0$