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

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:

