Question

What is the intersection of this sphere with the yz-plane? find an equation of the sphere with center (−3, 2, 9) and radius 6?

Answer 1

The equation of the sphere with center (-3, 2, 9) and radius 6 is (x + 3)² + (y - 2)² + (z - 9)² = 6². The intersection of this sphere with the yz-plane is described by the equation (y - 2)² + (z - 9)² = 27.

Step-by-step explanation:

The equation of a sphere with center (-3, 2, 9) and radius 6 is given by:

(x + 3)² + (y - 2)²+ (z - 9)² = 6²

To find the intersection of this sphere with the yz-plane, we need to set x = 0 in the equation:

(0 + 3)² + (y - 2)² + (z - 9)² = 6²

Simplifying this equation gives:

(y - 2)² + (z - 9)² = 27

So, the intersection of the sphere with the yz-plane is described by the equation (y - 2)² + (z - 9)² = 27.

Answer 2

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


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

so the equation of the sphere with center (-3, 2, 9) and radius 6 is :


`(x-(-3))^(2) + (y-2)^(2) + (z-9)^(2) = 6^(2)`


`(x+3)^(2) + (y-2)^(2) + (z-9)^(2) = 36`


The yz plane is the set of all points (0, y, z), that is x is always 0.

For x=0, the plane


`(x+3)^(2) + (y-2)^(2) + (z-9)^(2) = 36`

becomes


`(0+3)^(2) + (y-2)^(2) + (z-9)^(2) = 36`


`(y-2)^(2) + (z-9)^(2)=36-9=25= 5^(2)`


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

This is the circle with center (y, z)=(2, 9) and radius 5.