Question

Find an equation of a sphere if one of its diameters has endpoints (4, 1, 6) and (8, 3, 8).

Answer 1

Answer:

The equation of a sphere with endpoints at (4, 1, 6) and (8, 3, 8) is $(x-6)^{2}+(y-2)^{2}+(z-7)^{2} = 6$.

Step-by-step explanation:

Given the extremes of the diameter of the sphere, its center is the midpoint, whose location is presented below:

$C(x,y,z) = \left(\frac{4+8}{2},\frac{1+3}{2},\frac{6+8}{2}\right)$

$C(x,y,z) = (6,2,7)$

Any sphere with a radius $r$ and centered at $(h,k,s)$ is represented by the following equation:

$(x-h)^{2}+(y-k)^{2}+(z-s)^{2} = r^{2}$

Let be $(x,y,z) = (4,1,6)$ and $(h,k,s) = (6,2,7)$, the radius of the sphere is now calculated:

$(4-6)^{2}+(1-2)^{2}+(6-7)^{2}=r^{2}$

$r = \sqrt{6}$

The equation of a sphere with endpoints at (4, 1, 6) and (8, 3, 8) is $(x-6)^{2}+(y-2)^{2}+(z-7)^{2} = 6$.