# The Intersection Between a Plane and a Sphere

When a spherical surface and a plane intersect, the intersection is a point or a circle. Here, we will be taking a look at the case where it’s a circle. Go here to learn about intersection at a point.

More often than not, you will be asked to find the distance from the center of the sphere to the plane and the radius of the intersection. This is how you do that:

• Imagine a line from the center of the sphere, $C$, along the normal vector that belongs to the plane. This line will hit the plane in a point $A$. The length of the line segment between the center and the plane can be found by using the formula for distance between a point and a plane.

• You can imagine another line from the center to a point $B$ on the circle of intersection. The length of this line will be equal to the radius of the sphere.

• The line along the plane from $A$ to $B$ is as long as the radius of the circle of intersection.

• The three points $A$, $B$ and $C$ form a right triangle, where the angle between $\stackrel{\to }{CA}$ and $\stackrel{\to }{AB}$ is $90$°. You can use Pythagoras’ theorem on this triangle. That means you can find the radius of the circle of intersection by solving the equation

 $\phantom{\rule{-0.17em}{0ex}}{|\stackrel{\to }{AB}|}^{2}+\phantom{\rule{-0.17em}{0ex}}{|\stackrel{\to }{CA}|}^{2}=\phantom{\rule{-0.17em}{0ex}}{|\stackrel{\to }{CB}|}^{2}$

for $\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{AB}|$.

Example 1

You have a circle with radius $R=3$ and its center in $C=\phantom{\rule{-0.17em}{0ex}}\left(-2,1,0\right)$. Find the distance from $C$ to the plane

 $x-3y-2z-1=0,$

and find the radius $r$ of the circle of intersection.

First, you find the distance from the center to the plane by using the formula for the distance between a point and a plane. That gives you $\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{CA}|& =\frac{\phantom{\rule{-0.17em}{0ex}}|a{x}_{1}+b{y}_{1}+c{z}_{1}+d|}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\phantom{\rule{-0.17em}{0ex}}|\phantom{\rule{-0.17em}{0ex}}\left(-2\right)-3\cdot 1-2\cdot 0-1|}{\sqrt{1+\phantom{\rule{-0.17em}{0ex}}{\left(-3\right)}^{2}+\phantom{\rule{-0.17em}{0ex}}{\left(-2\right)}^{2}}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{6}{\sqrt{14}}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Using Pythagoras’ theorem, you get $\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}{|\stackrel{\to }{AB}|}^{2}+\phantom{\rule{-0.17em}{0ex}}{|\stackrel{\to }{CA}|}^{2}& =\phantom{\rule{-0.17em}{0ex}}{|\stackrel{\to }{CB}|}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {r}^{2}+\phantom{\rule{-0.17em}{0ex}}{\left(\frac{6}{\sqrt{14}}\right)}^{2}& ={3}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {r}^{2}& =9-\frac{36}{14}=\frac{45}{7}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill r& =\sqrt{\frac{45}{7}}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

You have found that the distance from the center of the sphere to the plane is $\frac{6}{\sqrt{14}}$, and that the radius of the circle of intersection is $\frac{\sqrt{45}}{\sqrt{7}}$.