Animated Solution for Mathematics - Three Dimensional Geometry: The plane $x+2y-z=4$ cuts the sphere $x^2+y^2+z^2-x+z-2=0$ in a circle of radius

Imagine you are holding a perfectly spherical orange. When you slice through it, the cross-section is a circle. This is the fundamental beauty of 3D geometry.

In this problem, we are given a sphere defined by the equation $x^2+y^2+z^2-x+z-2=0$ and a plane $x+2y-z=4$. Our mission is to find the radius of the circle formed by this intersection.

Unmasking the Sphere

To understand the sphere, we compare the given equation to the general form:

By comparing coefficients, we find $2u=-1$, $2v=0$, and $2w=1$. This yields the center $C=(-u,-v,-w)=(\frac{1}{2},0,-\frac{1}{2})$.

The radius $R$ is calculated using the formula $R=\sqrt{u^2+v^2+w^2-d}$. Substituting our values:

The Perpendicular Bridge

Next, we determine the perpendicular distance $p$ from the center of the sphere to the plane. We use the formula:

Substituting the center $C(\frac{1}{2},0,-\frac{1}{2})$ into the plane equation $x+2y-z-4=0$:

The Pythagorean Triumph

Visualize the right-angled triangle formed by the sphere's center $C$, the center of the circle $M$, and a point $P$ on the circle's circumference. The hypotenuse is the sphere's radius $R$, the base is the circle's radius $r$, and the height is the distance $p$.

By the Pythagorean theorem, $r^2+p^2=R^2$, which implies $r=\sqrt{R^2-p^2}$. Substituting our calculated values:

The radius of the circle is 1. You have conquered the geometry!