Question

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

Answer 1

Find the equation of sphere:

The general equation of a sphere with a center$C=(x_c,y_c,z_c)$ and radius $r$ is ${(x-x_c)}^2+{(y-y_c)}^2+{(z-z_c)}^2=r^2$

Here, the center is $C=(4,1,2)$ and sphere is passing through origin.

So the radius is the distance between point $C$ and the origin, which is given by;

$\Rightarrow$$r=\sqrt{{(x_c-x_o)}^2+{(y_c-y_o)}^2+{(z_c-z_o)}^2}$

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

$$\begin{gathered} \Rightarrow r=\sqrt{16+1+4} \\ \Rightarrow r=\sqrt{21}\mathrm{units} \end{gathered}$$

Now substitute the value of $r$ in ${(x-x_c)}^2+{(y-y_c)}^2+{(z-z_c)}^2=r^2$, we get;

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

Hence, the equation of the sphere is ${(x-4)}^2+{(y-1)}^2+{(z-2)}^2=21$.