Find the equation of the sphere that has the points $( 8,2,2 )$ and $( 6,4,4 )$ as end points of a diameter.
A) $( x + 7 ) + ( y + 3 ) + ( z + 3 ) = 6$
B) $( x + 7 ) ^ { 2 } + ( y + 3 ) ^ { 2 } + ( z + 3 ) ^ { 2 } = 3$
C) $( x - 7 ) ^ { 2 } + ( y - 3 ) ^ { 2 } + ( z - 3 ) ^ { 2 } = 6$
D) $( x - 7 ) ^ { 2 } + ( y - 3 ) ^ { 2 } + ( z - 3 ) ^ { 2 } = 3$
E) $( x - 7 ) + ( y - 3 ) + ( z - 3 ) = 3$

Question

Quizplus · Accepted Answer

The center of the sphere is the midpoint of the diameter, which is calculated as $\left( \frac{8+6}{2}, \frac{2+4}{2}, \frac{2+4}{2} \right) = (7, 3, 3)$. The radius is half the distance between the two points, which can be found using the distance formula: $\sqrt{(8-6)^2 + (2-4)^2 + (2-4)^2} = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{12}$. Half of this distance is $\sqrt{3}$, so the radius squared is 3. Thus, the equation of the sphere is $(x - 7)^2 + (y - 3)^2 + (z - 3)^2 = 3$.

Find the Equation of the Sphere That Has the Points (8,2,2)( 8,2,2 )(8,2,2)

Find the Equation of the Sphere That Has the Points $( 8,2,2 )$