How do I find the equation of a sphere that passes through the origin and whose center is $(4, 1, 2)$ $(4, 1, 2)$ ?

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Answer 1 · 2015-10-28T21:05:52

The equation is: ${(x - 4)}^{2} + {(y - 1)}^{2} + {(z - 2)}^{2} = 21$ $(x-4)^2+(y-1)^2+(z-2)^2=21$

The general equation of a sphere with a center $C = (x_{c}, y_{c}, z_{c})$ $C=(x_c,y_c,z_c)$ and radius $r$ $r$ is:

${(x - x_{c})}_{c}^{2} + {(y - y_{c})}_{c}^{2} + {(z - z_{c})}_{c}^{2} = r^{2}$ $(x-x_c)^2+(y-y_c)^2+(z-z_c)^2=r^2$

In this case center is given $C = (4, 1, 2)$ $C=(4,1,2)$ .

To calculate the radius we use the second point given - the origin. So the radius is the distance between point $C$ $C$ and the origin:

$r = \sqrt{{(x_{C} - x_{O})}_{C}^{2} + {(y_{C} - y_{O})}_{C}^{2} + {(z_{C} - z_{O})}_{C}^{2}} =$ $r=sqrt((x_C-x_O)^2+(y_C-y_O)^2+(z_C-z_O)^2)=$

$= \sqrt{{(4 - 0)}^{2} + {(1 - 0)}^{2} + {(2 - 0)}^{2}} = \sqrt{16 + 1 + 4} = \sqrt{21}$ $=sqrt((4-0)^2+(1-0)^2+(2-0)^2)=sqrt(16+1+4)=sqrt(21)$

Now when we have all required data we can write the equation of the sphere:

${(x - 4)}^{2} + {(y - 1)}^{2} + {(z - 2)}^{2} = 21$ $(x-4)^2+(y-1)^2+(z-2)^2=21$

How do I find the equation of a sphere that passes through the origin and whose center is (4,1,2)(4, 1, 2)?