Find the standard equation of the sphere?

Question

Find the standard equation of the sphere with center (-3, 2, 4) that is tangent to the plane given by
2x + 4y - 3z = 8.

Answer 1

The distance from the center of the sphere to the plane is the radius of the sphere. The point (4, 0, 0) is on the plane. We can take the vector initiating at (-3, 2, 4) and terminating at (4, 0, 0) and project this onto the normal to the plane. The length of the projection is the distance sought.



u = (4, 0, 0) - (-3, 2, 4) = (7, -2, -4).



The unit normal to the sphere is



n = (2, 4, -3)/√(2² + 4² + 3²) = (2, 4, -3)/√(29).



The projection of u onto n is



proj_n(u) = (u∙n)n = (14 - 8 + 12)/√(29) (2, 4, -3)/√(29) = (18/29)(2, 4, -3)



The length of this vector is



r = (18/29)√(29) = 18/√(29).



The sphere has equation



(x + 3)² + (y - 2)² + (z - 4)² = 18²/29

Answer 2

The closest point (x,y,z) of the plane to point (-3,2,4) satisfies both equation of the plane and the condition that difference (x,y,z)-(-3,2,4) is collinear with normal vector of the plane, which is (2,4,-3). So, we put x=-3+2a, y=2+4a, z=4-3a into plane's equation in order to calculate the difference.

2(-3+2a)+4(2+4a)-3(4-3a)=8

-6+4a+8+16a-12+9a=8

-18+29a=0

a=18/29

(x,y,z)-(-3,2,4)=18/29(2,4,-3)

Norm r=18sqrt(29)/29 of this difference is radius of the sphere, so its equation is

(x+3)^2+(y-2)^2+(z-4)^2=324/29

It remains to expand squares of binomials on the left-hand-side and simplify the equation.

Answer 3

4/3pie r 2