Question

Find an equation of the sphere with center (4, −12, 8) and radius 10.

Answer 1

Answer:

Step-by-step explanation:

To find an equation of the sphere with center (4, −12, 8) and radius 10

$(x-4)^2+(y+12)^2+(z-8)^2 = 100$

intersection with xy-plane

Put z=0

$(x-4)^2+(y+12)^2+(0-8)^2 = 100$

$(x-4)^2+(y+12)^2 = 36$

(A circle with centre at (4,-12) and radius 6)

intersection with xz-plane

Put y =0

$(x-4)^2+(0+12)^2+(z-8)^2 = 100$

$(x-4)^2+(z-8)^2 = -44$

Sum of squares cannot be positive, so DNE

intersection with yz-plane

Put x=0

$(0-4)^2+(y+12)^2+(z-8)^2 = 100$

$(y+12)^2+(z-8)^2 = 84$

A circle in YZ plane with centre at y =-12 and z =8 and radius square root of 84