Tangent Line From a Point on a Sphere and $y$-axis

Question

Let's say I have a sphere,

$$100 = x^2+y^2 +z^2 $$

This indicates that the center of our sphere is at $$(0, 0, 0)$$ and we have a radius of $$radius = 10$$

I'm under the assumption that $$P = (1, 9, \sqrt{18})$$ is a point on the sphere, correct me if I'm wrong.

Let's say I have a point $P$ and $3$ axes -- how would I calculate the tangent line that passes through the $p$ and the $y$-axis?

Answer 1

Approach: Intersection of tangent plane to the sphere at P with Y axis, call it Q. Line that contains PQ.

The equation of the tangent plane is:

$$x+9y+\sqrt{18} z+D=0$$

And $$1+81+18+D=0$$

The plane is then $$x+9y+\sqrt{18} z-100=0$$

And Q is $(0, 100/9,0)$

PQ is then $( -1,-9+100/9,-\sqrt{18})$

And the desired line is

$$P + \lambda PQ$$

That is to say:

\begin{align} x&= 1-\lambda \nonumber \\ y&= 9 + \frac{19}{9} \lambda. \nonumber \\ z&= \sqrt{18} -\sqrt{18} \lambda. \nonumber \end{align}

which is the equation of the line in parametric coordinates.

In cartesian coordinates:

$$1-x=\frac{9}{19}(y-9) = \frac{\sqrt{18}-z}{\sqrt{18}}$$

Pictorially:

Answer 2

Given C(0,0,0) the center of the sphere, A(1,9,$\sqrt {18}$) a point of the sphere and R(0,y1,0) the point of intersection of the line with the Y axis. We can write a generic vectorial equation for the tangent line:

(x,y,z) = A + tk, k∈Z in which t is a directional vector of the tangent line

Since the tangent line is perpendicular to the sphere on A we have the vector: AC = A - C = (1,9,$\sqrt {18}$) that is perpendicular to the line t, which allow us to write the plane x + 9y + 9$\sqrt {18}$z = d that has the tangent line, the point A and the point R.

Since A belongs to the plane:

1+9*9+$\sqrt {18}$ * $\sqrt {18}$ = d

<=> 1 + 81 + 18 = d

<=> 100 = d

The the equation of the plane is: x +9y + $\sqrt {18}$z = 100.

Since R belongs to the plane:

0 + 9y1 + 0 = 100

<=> 9y1 = 100

<=> y1 = $\frac{100}{9}$

Therefore R(0,$\frac{100}{9}$,0)

Now we can find the directional vector t using R and A (two points of the tangent line)

RA = A - R = (1, $\frac{-19}{9}$,$\sqrt{18}$)

Finally, having a point of the tangent line (point A or point R) and a vector (vector RA) we can find the vectorial equation of the tangent line: (x,y,z) = (1,9,$\sqrt{18}$) + (1, $\frac{-19}{9}$,$\sqrt{18}$)k, k∈Z

You could also find the cartesian equation of the line:

(x,y,z) = (1,9,$\sqrt{18}$) + (1, $\frac{-19}{9}$,$\sqrt{18}$)k, k∈Z

<=> (x,y,z) = (1+k,9+$\frac{-19}{9}$k,$\sqrt{18}$+$\sqrt{18}$k), k∈Z

<=> $\begin{cases} x = 1+k \\ y = 9+\frac{-19}{9}k \\ z = \sqrt{18}+\sqrt{18}k \end{cases} $ <=> $\begin{cases} x-1 = k \\ \frac{y-9}{\frac{-19}{9}} = k \\ \frac{z-\sqrt{18}}{\sqrt{18}} = k \end{cases} $

<=> x-1 = 9*$\frac{9-y}{19}$ = $\frac{z-\sqrt{18}}{\sqrt{18}}$

Answer 3

These lines can be constructed without explicitly constructing the tangent plane or computing its intersections with the coordinate axes. If you look straight down at the tangent plane, the coordinate axes will appear to be concurrent lines that pass through the given point. These projections of the coordinate axes are exactly the lines you seek.

You already know a point on each of these lines ($p$), so you just need to find direction vectors $d$ for them with which to construct a parameterization $p+td$ of the line, but those direction vectors are just the orthogonal projections of the unit coordinate vectors onto the tangent plane. These are equal to the orthogonal rejections from the tangent plane’s normal, which for a sphere is equal to the displacement vector of the point of tangency, and so, using a standard formula for projection, the three direction vectors are $$(1,0,0)-{p_x\over p\cdot p}p \\ (0,1,0)-{p_y\over p\cdot p}p \\ (0,0,1)-{p_z\over p\cdot p}p.$$