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

Oct 28, 2015

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

#### Explanation:

The general equation of a sphere with a center $C = \left({x}_{c} , {y}_{c} , {z}_{c}\right)$ and radius $r$ is:

${\left(x - {x}_{c}\right)}^{2} + {\left(y - {y}_{c}\right)}^{2} + {\left(z - {z}_{c}\right)}^{2} = {r}^{2}$

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

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

$r = \sqrt{{\left({x}_{C} - {x}_{O}\right)}^{2} + {\left({y}_{C} - {y}_{O}\right)}^{2} + {\left({z}_{C} - {z}_{O}\right)}^{2}} =$

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

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

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