# How to find the intersection of a triangle with a circle

I'm trying to calculate the penumbra of the upcoming eclipse, which involves finding the inner tangents of the Sun to the Moon and the seeing how those rays project on the Earth:

I made a simulation in Mathematica with grossly inaccurate radii just to work out the calculations, and got most of the way there. (The Sun is orange, the Moon gray, the earth Green. Yes, I know the Earth isn't larger than the Sun! Just picking radii that fit nicely on the screen.)

Where I'm getting hung up is finding the coordinates of the two points circled in red. I know the distance from the internal homothetic center to the edge of the Earth -- the straight red line -- as was as the angle from the inner homothetic center relative to the plane (B) and, of course, the radius of the fake Earth.

But for the life of me, I can't figure out how the inner tangent lines intersect with the radius of the Earth (the big green circle). It would be easy if the Earth was flat, but I'm led to understand that's not the case :)

EDIT:

I was able to solve this by brute algebra, converting the Earth to x^2 + y^2 = r^2 and finding the intersection with the line:

I was hoping for a geometric solution, but this works!

• Why don't you just put the whole thing in coordinate system, calculate common tangents of Sun and Moon, then find intersection of line and circle, which is a system of second and first order equation, which boils down to quadratic equation? Commented Jul 29, 2017 at 16:40
• Is my update what you had in mind? Commented Jul 29, 2017 at 16:50
• Yes, something like that. Admittedly, it's uglier than I thought it'd be. Commented Jul 29, 2017 at 17:18

In the figure:

from the configuartion of the eclipse we know:

$\overline{AE}=r$

$\overline{AD}=l$

$\angle EDA=\alpha$

and we can find $\beta= \angle AED$ using the sin rule: $$\frac{r}{\sin \alpha}=\frac{l}{\sin\beta}$$

Now we can find $\angle EAD=\pi-\alpha-\beta$ and , from this we find $EF$ and/or the arc $EG$.

• Worked brilliantly! The only hiccup was that the law of Sines was giving me the outer point, to the left of the center, but I fixed that with Pi - B. Thanks! Commented Jul 29, 2017 at 20:26

If $H$ is the internal homothetic center, $O$ the center of the Earth (considered as a sphere) and $A$ one of the two points you need, you can apply sine rule to triangle $HOA$ to find $\angle AOH$, remembering that $\angle OAH=\pi-(\angle HOA+\angle OHA)$: $${\sin(\angle HOA+\angle OHA)\over OH}={\sin(\angle OHA)\over OA}.$$

• See my edit -- I reverted to algebra though I was hoping for a cleaner geometric solution. I'm afraid I'm not quite numerate enough to follow this solution... Commented Jul 29, 2017 at 16:51