# Intersection Between a Tangent Plane and a Sphere

When a sphere and a plane intersects, the intersection can be described as either a point or a circle. Here, we will learn about the case when it’s a single a point. Go here to learn about intersection as a circle.

When the intersection between a tangent plane and a sphere surface is a point, we call this point the tangent point. To find this point, you do the following:

You can see that the vector extending from the center of the sphere to the tangent point have to be perpendicular to the tangent plane, because the angle between a tangent and the line from the tangent point to the center of the sphere is always $90$°. That means this vector is parallel to the normal vector to the tangent plane, and you can create a line through the tangent point and the center of the sphere with that normal vector as a directional vector. To find the tangent point, you have to find the intersection between this line and the tangent plane.

Example 1

You have a spherical surface that has its center in $\phantom{\rule{-0.17em}{0ex}}\left(1,1,2\right)$ and a tangent plane given by the equation

 $x+2y=0.$

Find the tangent point.

You can see that $\phantom{\rule{-0.17em}{0ex}}\left(1,2,0\right)$ is a normal vector to the plane. That means the parametric equation for the line through the center with the normal vector as its directional vector is

 $x\phantom{\rule{-0.17em}{0ex}}\left(t\right)=1+t,\phantom{\rule{0.33em}{0ex}}y\phantom{\rule{-0.17em}{0ex}}\left(t\right)=1+2t,\phantom{\rule{0.33em}{0ex}}z\phantom{\rule{-0.17em}{0ex}}\left(t\right)=2.$

 $x\phantom{\rule{-0.17em}{0ex}}\left(t\right)=1+t,\phantom{\rule{2em}{0ex}}y\phantom{\rule{-0.17em}{0ex}}\left(t\right)=1+2t,\phantom{\rule{2em}{0ex}}z\phantom{\rule{-0.17em}{0ex}}\left(t\right)=2.$

Insert this into the equation for the tangent plane. That gives you $\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}\left(1+t\right)+2\phantom{\rule{-0.17em}{0ex}}\left(1+2t\right)& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 3+5t& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill t& =-\frac{3}{5}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

You insert for $t$ in the parametric equation and get

 $\phantom{\rule{-0.17em}{0ex}}\left(1+\frac{-3}{5},1+2\frac{-3}{5},2\right)=\phantom{\rule{-0.17em}{0ex}}\left(\frac{2}{5},-\frac{1}{5},2\right).$

This point will be the intersection between the line through the center and the tangent plane, which also makes it the tangent point.

Formula

### TangentPointofSphericalSurfaceandPlane

The tangent point $T$ where a plane with a normal vector $\stackrel{\to }{n}$ intersects a sphere with radius $r$ and its center at $C$ is given by

 $\stackrel{\to }{OT}=\stackrel{\to }{OC}+r\cdot \frac{1}{\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{n}|}\cdot \stackrel{\to }{n}.$

Example 2

You will often be asked to show that a plane is tangent to a spherical surface and to find the tangent point. A plane is tangent to a spherical surface if the distance from the center $c$ of the sphere to the plane $\alpha$ is equal to the radius of the sphere.

Say you have the sphere

 $\phantom{\rule{-0.17em}{0ex}}{\left(x-2\right)}^{2}+\phantom{\rule{-0.17em}{0ex}}{\left(y-1\right)}^{2}+{z}^{2}=9$

and the plane

 $2x+y-2z+4=0.$

From the equation of the sphere, you can see that the radius is $r=\sqrt{9}=3$ and the center is at $\phantom{\rule{-0.17em}{0ex}}\left(2,1,0\right)$. To show that the plane is tangent to the surface of the sphere, you use the formula for the distance between a point and a plane and check whether the answer is equal to the radius of the sphere, which we know is $3$: $\begin{array}{llll}\hfill \frac{\phantom{\rule{-0.17em}{0ex}}|2\phantom{\rule{-0.17em}{0ex}}\left(2\right)+1\cdot 1-2\phantom{\rule{-0.17em}{0ex}}\left(0\right)+4|}{\sqrt{{2}^{2}+{1}^{2}+\phantom{\rule{-0.17em}{0ex}}{\left(-2\right)}^{2}}}& =\frac{\phantom{\rule{-0.17em}{0ex}}|4+1+0+4|}{\sqrt{9}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\phantom{\rule{-0.17em}{0ex}}|9|}{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

To find the tangent point where the surface of the sphere and the plane touches, you can do like in Example 1. You can also use the formula above for the tangent point of a spherical surface and a plane.

Note! Take a note of that in this case, the normal vector can send you in the wrong direction. Then you will end up at the opposite side of the sphere. For that reason, check whether your answer lies in the tangent plane. If it doesn’t, simply switch $\stackrel{\to }{n}$ with $-\stackrel{\to }{n}$.

In this case, the radius of the sphere and the distance between the center of the sphere and the plane are the same, giving you that

 $\phantom{\rule{-0.17em}{0ex}}\left(2,1,0\right)+3\frac{1}{3}\phantom{\rule{-0.17em}{0ex}}\left(2,1,-2\right)=\phantom{\rule{-0.17em}{0ex}}\left(4,2,-2\right).$

Put this into the equation of the tangent plane to check whether you have gone in the right direction. That gives you

$\begin{array}{llll}\hfill 2x+y-2z+4& =2\cdot 4+1\cdot 2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}-2\cdot \phantom{\rule{-0.17em}{0ex}}\left(-2\right)+4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =8+2+4+4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =18\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \ne 0.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill 2x+y-2z+4& =2\cdot 4+1\cdot 2-2\cdot \phantom{\rule{-0.17em}{0ex}}\left(-2\right)+4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =8+2+4+4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =18\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \ne 0.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

You can see that this does not satisfy the equation of the plane, which means the normal vector has sent you in the wrong direction. If you change the sign in front of the normal vector, you get
 $\phantom{\rule{-0.17em}{0ex}}\left(2,1,0\right)-3\frac{1}{3}\phantom{\rule{-0.17em}{0ex}}\left(2,1,-2\right)=\phantom{\rule{-0.17em}{0ex}}\left(0,0,2\right).$

Put this into the equation of the plane, and you get

$\begin{array}{llll}\hfill 2x+y-2z+4& =2\cdot 0+0-2\cdot 2+4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-4+4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $2x+y-2z+4=2\cdot 0+0-2\cdot 2+4=-4+4=0.$

Then, $\stackrel{\to }{OT}=\phantom{\rule{-0.17em}{0ex}}\left(0,0,2\right)$.