Given a sphere with the equation of $x^2+y^2+z^2=144$ and the plane $x=5$, find the equation of the resulting intersection.
I know that the intersection will be a circle. I tried to plug x in as five. However, when graphing this online, I realized that this was a cylinder. I am not sure how to limit this region to only the intersection.
Here $y^2+z^2=119$ would indeed be a cylinder, if $x$ can take any value,
but that is not an expression for this intersection, since we know $x=5$,
so the answer is "$y^2+z^2=119$ and $x=5$", which is a circle.
You could make this a messy single equation such as $(y^2+z^2-119)^2 +(x-5)^2=0$ though it would not be particularly informative and not so obviously be a circle