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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?

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4 Answers 4

up vote 2 down vote accepted

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.

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Awesome Thanks a million! –  Cactus BAMF Sep 22 '12 at 19:38

Note that, the point of tangent is projection of the center $I(\alpha, \beta, \gamma)$ of the sphere $(S)$ onto the plane $(P)$. Therefore, you can do following steps.

1) Write the equation of the line $\Delta$ passing the point $I$ and perpendicular to the plane $(P)$, Parallel vetor of $\Delta$ is also Normal vector of the plane $(P)$.

2) The coordinates of the point of tangent is solution of the system of two equations: Equation of the plane $(P)$ and equation of the line $\Delta$.

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The distance between point C $(\alpha, \beta, \gamma)$ and plane $ax+by+cz-d=0$ is $R=\frac{a\alpha+b\beta+c\gamma-d}{\sqrt{a^2+b^2+c^2}}$

But the distance$(R)$ between centre and tangent plane is the radius$(r)$ of the sphere, so the equation is $(x−α)^2+(y−β)^2+(z−γ)^2=\frac{(a\alpha+b\beta+c\gamma-d)^2}{a^2+b^2+c^2}$

Regarding the intersection, one may look into this for the general idea.

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The normal to the plane is $ai + bj + ck$. Write and equation for the line going through the prescribed point and tell where it hits the plane. Let $r$ be the distance between the point and the plane (along the perpendicular line). Your equation will b3 $$(x - \alpha)^2 + (y - \beta)^2 + (z - \gamma)^2 = r^2.$$

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