Find the standard form of the equation of the sphere with the given characteristics. Endpoints of a diameter: (-1,6,1), (7,-4,-9)
A) $( x - 3 ) ^ { 2 } + ( y - 1 ) ^ { 2 } + ( z - 4 ) ^ { 2 } = 66$
B) $( x + 3 ) ^ { 2 } + ( y - 1 ) ^ { 2 } + ( z + 4 ) ^ { 2 } = 66$
C) $( x - 3 ) ^ { 2 } + ( y + 1 ) ^ { 2 } + ( z + 4 ) ^ { 2 } = 264$
D) $( x - 3 ) ^ { 2 } + ( y - 1 ) ^ { 2 } + ( z - 4 ) ^ { 2 } = 132$
E) $( x - 3 ) ^ { 2 } + ( y - 1 ) ^ { 2 } + ( z + 4 ) ^ { 2 } = 66$

Question

Quizplus · Accepted Answer

The center of the sphere is the midpoint of the diameter, which is calculated as $\left( \frac{-1+7}{2}, \frac{6-4}{2}, \frac{1-9}{2} \right) = (3, 1, -4)$. The radius is half the distance between the endpoints, which can be found using the distance formula: $\sqrt{(7-(-1))^2 + (-4-6)^2 + (-9-1)^2} = \sqrt{8^2 + (-10)^2 + (-10)^2} = \sqrt{64 + 100 + 100} = \sqrt{264}$. The radius of the sphere is half of this, so $r = \sqrt{66}$. Therefore, the equation of the sphere is $(x - 3)^2 + (y - 1)^2 + (z + 4)^2 = 66$.

Find the Standard Form of the Equation of the Sphere (x−3)2+(y−1)2+(z−4)2=66( x - 3 ) ^ { 2 } + ( y - 1 ) ^ { 2 } + ( z - 4 ) ^ { 2 } = 66(x−3)2+(y−1)2+(z−4)2=66

Find the Standard Form of the Equation of the Sphere $( x - 3 ) ^ { 2 } + ( y - 1 ) ^ { 2 } + ( z - 4 ) ^ { 2 } = 66$