Answer to Question #165289 in Calculus for Alyssa

Question

Find the equation of the sphere with the given conditions:

(a) Center (2,4,3), tangent plane 2x-3y+4z-5=0

(b) endpoints of the diameter are (-5,3,2) and (-1,-3,2)

Answer 1

(a) Let us find the equation of the sphere with the center $(2,4,3)$ and tangent plane $2x-3y+4z-5=0$. The radius of the sphere is equal to the distance from the center to the tangent plane: $R=\frac{|2\cdot 2-3\cdot 4+4\cdot 3-5|}{\sqrt{2^2+(-3)^2+4^2}}=\frac{1}{\sqrt{29}}.$ Consequently, the equation of the sphere is the following:

$(x-2)^2+(y-4)^2+(z-3)^2=\frac{1}{29}$

(b) Let us find the equation of the sphere in the case that the endpoints of the diameter are $(-5,3,2)$ and $(-1,-3,2)$ . It follows that the center of the sphere is $O(a,b,c)$, where $a=\frac{-5-1}{2}=-3,\ b=\frac{3-3}{2}=0,\ c=\frac{2+2}{2}=2.$ The radius is equal the distance from the center to $(-1,-3,2):$ $R=\sqrt{(-1+3)^2+(-3-0)^2+(2-2)^2}=\sqrt{13}$. Therefore, the equation of the sphere is the following:

$(x+3)^2+y^2+(z-2)^2=13$