Question

Find an equation of the sphere with center (-3, 2 , 5) and radius 4. What is the intersection of this sphere with the yz-plane.

Answer 1

The equation of the sphere with center (-3, 2 , 5) and radius 4 is
`(x+3)^(2) +(y-2)^(2) + (z-5)^(2) = 16`

The intersection of the sphere with the yz- plane gave the equation
`(y-2)^(2) + (z-5)^(2) = 7` which is a 2D- circle with center (0,2,5) and radius
`√(7)`.

Explanation:

The equation of a sphere of radius r, with center (a,b,c) is given by

`(x-a)^(2) +(y-b)^(2) + (z-c)^(2) = r^(2)`

where,
`x,`
`y,` and
`z` are the coordinates of the points on the surface of the sphere.

Hence, the equation of the sphere with center, (-3, 2 , 5) and radius 4 becomes

`(x-a)^(2) +(y-b)^(2) + (z-c)^(2) = r^(2)`

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

Then,

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

This is the equation of the sphere with center (-3, 2 , 5) and radius 4,

Now, for the intersection of this sphere with the yz- plane,

The
`yz -`plane is where
`x = 0`, then we set
`x = 0`

Them the equation
`(x+3)^(2) +(y-2)^(2) + (z-5)^(2) = 16` becomes

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

`(3)^(2) +(y-2)^(2) + (z-5)^(2) = 16\\9 +(y-2)^(2) + (z-5)^(2) = 16\\(y-2)^(2) + (z-5)^(2) = 16 - 9\\(y-2)^(2) + (z-5)^(2) = 7`

This equation is the equation of a 2D- circle with center (0,2,5) and radius
`√(7)`

This is the part of the sphere that intersects with the yz-plane.