Question

Find an equation of the sphere with center (−2, 3, 7) and radius 7.What is the intersection of this sphere with the yz-plane?

Answer 1

Answer:

The equation of the sphere with center (-2, 3, 7) and radius 7 is $(x+2)^{2}+(y-3)^{2}+(z-7)^{2} = 49$.

The intersection of the sphere with the yz-plane is $(y-3)^{2}+(z-7)^{2} = 49$.

Step-by-step explanation:

We know that any sphere can be represented by the following equation:

$(x-h)^{2}+(y-k)^{2}+(z-s)^{2} = r^{2}$

Where:

$h$, $k$, $s$ - Coordinates of the center of the sphere, dimensionless.

$r$ - Radius of the sphere, dimensionless.

If $(h,k, s) = (-2,3,7)$ and $r = 7$, we obtain this expression:

$(x+2)^{2}+(y-3)^{2}+(z-7)^{2} = 49$

The intersection of the sphere with the yz-plane observe the following conditions:

$x = 0$, $y \in \mathbb {R}$, $z\in \mathbb{R}$

Hence, the expression above can be reduced into this:

$(y-3)^{2}+(z-7)^{2} = 49$