Finding the intersection of an xy-plane in a 3D-Coordinate System

Question

I found the equation of a sphere that has a center of $(1,-12,8)$ with a radius of 10 and I got the following equation:

$(x-1)^2 + (y+12)^2 + (z-8)^2 = 100$

As for finding an intersection for the xy-plane I got this and it was the wrong answer: $(x-1)^2 + (y+12)^2 = 100$

I assumed that it would be this because since it was an xy-plane, it wouldn't include the z-portion of the equation.

Answer 1

What your answer amounts to is the circle at which the sphere intersects the plane $z=8$.

However, we're looking for the intersection of the sphere and the $x$-$y$ plane, given by $z=0.\;$ In other words, we're looking for all points of the sphere at which the $z$-component is $0$.

Thus we need to evaluate the sphere using $z = 0,\,$ which yields the circle $$\begin{align} (x-1)^2 + (y+12)^2 + (0-8)^2 = 100 &\iff (x-1)^2 + (y+12)^2 + 64 = 100\\ \\ &\iff (x-1)^2 + (y+12)^2 = 36 = 6^2\end{align}$$

Thus the intersection of the sphere and the $x$-$y$ plane gives a circle with center $(1, -12)$ and radius $6$.

Answer 2

In the $xy$-plane, $z=0$, but note that you are looking at $(z-8)^2$ as one of the terms of your sphere.