Question

Find an equation of the sphere with center s23, 2, 5d and radius 4. What is the intersection of this sphere with the yz-plane?

Answer 1

Equation:
`(x+3)^2+(y-2)^2+(z-5)^2=16`

Intersection:
`(y-2)^2+(z-5)^2=7`

Step-by-step explanation:

We are asked to write an equation of the sphere with center center
`(-3,2,5)` and radius 4.

We know that equation of a sphere with radius 'r' and center at
`(h,k,l)` is in form:

`(x-h)^2+(y-k)^2+(z-l)^2=r^2`

Since center of the given sphere is at point
`(-3,2,5)`, so we will substitute
`h=-3`,
`k=2`,
`l=5` and
`r=4` in above equation as:

`(x-(-3))^2+(y-2)^2+(z-5)^2=4^2`

`(x+3)^2+(y-2)^2+(z-5)^2=16`

Therefore, our required equation would be
`(x+3)^2+(y-2)^2+(z-5)^2=16`.

To find the intersection of our sphere with the yz-plane, we will substitute
`x=0` in our equation as:

`(0+3)^2+(y-2)^2+(z-5)^2=16`

`9+(y-2)^2+(z-5)^2=16`

`9-9+(y-2)^2+(z-5)^2=16-9`

`(y-2)^2+(z-5)^2=7`

Therefore, the intersection of the given sphere with the yz-plane would be
`(y-2)^2+(z-5)^2=7`.