Question

At every point on a sphere $(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=r^2$ there is some plane tangent to the sphere. Explain how to find the equation of the tangent plane at any given point.

Short answer

$(\alpha -a)(x-\alpha )+(\beta -b)(y-\beta )+(\gamma -c)(z-\gamma )=0$

Step-by-step solution

Given information

The sphere $(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=r^2$and the point $(\alpha ,\beta ,\gamma )$

Calculation

The goal is to determine the equation of the plane tangent to the sphere at a specific position. The equation of the plane tangent to the sphere is obtained by finding the normal of the surface of the sphere $(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=r^2$at the given point $(\alpha ,\beta ,\gamma )$

The normal of the sphere $(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=r^2$at the point $(\alpha ,\beta ,\gamma )$is:

$$\begin{gathered} {\left.\nabla f(x,y,z)\right|}_{(\alpha ,\beta ,y)}=\left(2\right(x-a)\mathbf{i}+2(y-b)\mathbf{j}+2(y-c)\mathbf{k}{)}_{(a,\beta ,y)} \\ =(2(\alpha -a)\mathbf{i}+2(\beta -b)\mathbf{j}+2(\gamma -c)\mathbf{k}) \\ =⟨2(\alpha -a),2(\beta -b),2(\gamma -c)⟩ \\ =⟨(\alpha -a),(\beta -b),(\gamma -c)⟩ \end{gathered}$$

Calculation

The equation of the plane tangent to the sphere $(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=r^2$and the point $(\alpha ,\beta ,\gamma )$is:

$(\alpha -a)(x-\alpha )+(\beta -b)(y-\beta )+(\gamma -c)(z-\gamma )=0$

The equation of the plane tangent to the sphere $(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=r^2$and at the point $(\alpha ,\beta ,\gamma )$is $(\alpha -a)(x-\alpha )+(\beta -b)(y-\beta )+(\gamma -c)(z-\gamma )=0$