## Condition For Two Surfaces With a point in Common to Have a Common Tangent Plane

Suppose two surfaces
$F(x,y,z)=0, \; G(x,y,z)=0$
have a point
$(x_1,y_1,z_1)$
in common. What are the conditions on the partial derivatives for the surcease to have a common tangent?
The tangent plane to
$F(x,y,z)=0$
at the point
$(x_1,y_1,z_1)$
is
$\frac{ \partial F}{\partial x}|_{(x_1,y_1,z_1)} (x-x_1)+ \frac{ \partial F}{\partial y}|_{(x_1,y_1,z_1)} (y-y_1)+ \frac{ \partial F}{\partial z}|_{(x_1,y_1,z_1)} (z-z_1)=0$
.
The tangent plane to
$G(x,y,z)=0$
at the point
$(x_1,y_1,z_1)$
is
$\frac{ \partial G}{\partial x}|_{(x_1,y_1,z_1)} (x-x_1)+ \frac{ \partial G}{\partial y}|_{(x_1,y_1,z_1)} (y-y_1)+ \frac{ \partial G}{\partial z}|_{(x_1,y_1,z_1)} (z-z_1)=0$
.
For these two planes to be the same plane the partial differentials mus be a multiple of each other, so that
$\frac{\partial F}{\partial x}|_{(x_1,y_1,z_1)}=k \frac{\partial G}{\partial x}|_{(x_1,y_1,z_1)}, \; \frac{\partial F}{\partial y}|_{(x_1,y_1,z_1)}=k \frac{\partial G}{\partial y}|_{(x_1,y_1,z_1)}, \; \frac{\partial F}{\partial z}|_{(x_1,y_1,z_1)}=k \frac{\partial G}{\partial z}|_{(x_1,y_1,z_1)}$

If the tangent planes are orthogonal, then the dot product of the gradient vectors to the surface must be zero, so
\begin{aligned} \frac{\partial F}{\partial x}|_{(x_1,y_1,z_1)} \frac{\partial g}{\partial x}|_{(x_1,y_1,z_1)} &+ \frac{\partial F}{\partial y}|_{(x_1,y_1,z_1)} \frac{\partial G}{\partial y}|_{(x_1,y_1,z_1)} \\ &+ \frac{\partial F}{\partial z}|_{(x_1,y_1,z_1)} \frac{\partial G}{\partial z}|_{(x_1,y_1,z_1)}\\ &= (\nabla F) \cdot (\nabla G)=0 \end{aligned}
.