Question

Find the equation of the sphere passing through pleft parenthesis negative 2 comma 3 comma 4 right parenthesis(−2, 3, 4) and qleft parenthesis 8 comma negative 1 comma 5 right parenthesis(8, −1, 5) with its center at the midpoint of pq.

Answer 1

In  this question , the given points are

$P(-2,3,4), Q(8,-1,5)$

Center is at the midpoint of PQ . And to find the midpoint, we will use the following formula

$( \frac{x_{1}+x_{2}}{2}, \frac{y_{1} +y_{2}}{2} , \frac{z_{1}+z_{2}}{2} )$

Substituting the points in the formula, we will get

$(\frac{-2+8}{2}, \frac{3-1}{2}, \frac{4+5}{2} )= (3,1,4.5)$

And PQ gives the diameter, and to find the length of diameter, we will use the following formula

$d = \sqrt{ (x_{2} -x_{1})^2 + (y_{2} -y_{1})^2 + (z_{2} - z_{1})^2}$

Substituting the given values, we will get

$d = \sqrt{ (8+2)^2 + (-1-3)^2 +  (5-4)^2} = \sqrt{100 + 16+1} = \sqrt{117} = 3 \sqrt{13}$

Radius is half of diameter, that is

$r = \frac{3 \sqrt{13}}{2}$

Now we use the equation of sphere, which is

$(x-x_{0})^2 + (y-y_{0})^2 + (z-z_{0})^2 = r^2$

$(x-3)^2 + (y-1)^2 + (z-4.5)^2 = ( \frac{3 \sqrt{13}}{2} )^2 \\ (x-3)^2 + (y-1)^2 + (z-4.5)^2 = \frac{117}{4}  $