Find the equation for a sphere with center $(\alpha,\beta,\gamma)$ tangent to the plane $ax + by + cz = d$.
The sphere is $(x-\alpha)^2 + (y-\beta)^2 +(z-\gamma)^2 = r^2$ and I understand that some vector on the plane is orthogonal to the radius vector, but How do I find the point on the plane that is the tangent point?
Best Answer
Let $C = (\alpha, \beta, \gamma)$, $N=(a,b,c)$ and $P$ be the tangent point on the plane. Since the plane is tangent to the sphere, the line from $P$ to $C$ is orthogonal to the plane, hence it is a multiple of the normal.
So we have $C-P = r \frac{N}{\|N\|}$ (There is no need to normalize the normal :-), but it lets us interpret the constant $r$ as a radius, with the possible annoyance that it may be negative). Since $P= C-r \frac{N}{\|N\|}$ lies on the plane, we have $\langle P, N \rangle = d$, which gives $r = \frac{\langle C, N \rangle -d}{\|N\|}$, hence $P = C-\frac{\langle C, N \rangle -d}{\|N\|} \frac{N}{\|N\|}$.
So, explicitly, $$P = (\alpha, \beta, \gamma) - \frac{a \alpha + b \beta + c \gamma -d}{a^2+b^2+c^2} (a,b,c).$$
Or, of course, you could compute $r = \frac{a \alpha + b \beta + c \gamma -d}{\sqrt{a^2+b^2+c^2}}$ and then $P = C -r \frac{N}{\|N\|}$, but realize that $r$ may be negative.