Question

Find an equation of the sphere through the point (4,3,-1) and has center (3,8,1).

Answer 1

Answer:

$(x - 3)^{2} + (y - 8)^{2} + (z - 1)^{2} = 30$

Step-by-step explanation:

The general equation of a sphere is as follows:

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

In which the center is $(x_{c}, y_{c}, z_{c}$, and r is the radius.

In this problem, we have that:

$x_{c} = 3, y_{c} = 8, z_{c} = 1$

So

$(x - 3)^{2} + (y - 8)^{2} + (z - 1)^{2} = r^{2}$

through the point (4,3,-1)

This leads us to find the radius.

$(4 - 3)^{2} + (3 - 8)^{2} + (-1 - 1)^{2} = r^{2}$

$r^{2} = 1 + 25 + 4$

$r^{2} = 30$

So the equation of the sphere is:

$(x - 3)^{2} + (y - 8)^{2} + (z - 1)^{2} = 30$