Find the standard equation of a sphere that has diameter with the end points given below. $( 3 , - 2,4 ) , ( 7,12,4 )$
A) $( x - 5 ) ^ { 2 } + ( y - 5 ) ^ { 2 } + ( z - 4 ) ^ { 2 } = 53$
B) $( x - 5 ) ^ { 2 } + ( y - 5 ) ^ { 2 } + ( z - 4 ) ^ { 2 } = 9$
C) $( x + 5 ) ^ { 2 } + ( y - 5 ) ^ { 2 } + ( z - 4 ) ^ { 2 } = 53$
D) $( x - 5 ) ^ { 2 } + ( y + 5 ) ^ { 2 } + ( z + 4 ) ^ { 2 } = 53$
E) $( x - 5 ) ^ { 2 } + ( y + 5 ) ^ { 2 } + ( z - 4 ) ^ { 2 } = 53$

Question

Quizplus · Accepted Answer

The center of the sphere is the midpoint of the diameter, which is calculated as $\left( \frac{3+7}{2}, \frac{-2+12}{2}, \frac{4+4}{2} \right) = (5, 5, 4)$. The radius is half the distance between the two points, which can be found using the distance formula: $\sqrt{(7-3)^2 + (12-(-2))^2 + (4-4)^2} = \sqrt{4^2 + 14^2 + 0^2} = \sqrt{16 + 196} = \sqrt{212}$. The radius of the sphere is half of this distance, so $r = \sqrt{212} / 2$, but for the equation, we use $r^2 = 212/4 = 53$. Therefore, the standard equation of the sphere is $(x - 5)^2 + (y - 5)^2 + (z - 4)^2 = 53$.

Find the Standard Equation of a Sphere That Has Diameter (3,−2,4),(7,12,4)( 3 , - 2,4 ) , ( 7,12,4 )(3,−2,4),(7,12,4)

Find the Standard Equation of a Sphere That Has Diameter $( 3 , - 2,4 ) , ( 7,12,4 )$