Finding the intersection points of straight lines (from the intersection of outer circle radius and inner circle) and the outer circle

Question

Given two circles with the same origin, and:

• their radii are $r_1$ and $r_2$
• radii difference is $s=|r_2-r_1|$
• $(x_0,y_0)$ is an arbitrary point on the inner circle (that is, on the one with smaller radius, though it's also possible that $r_1=r_2$)

Extending two straight lines (i.e. one with no slope and one with slope of $0$) from $(x_0,y_0)$ find where these intersect the outer circle. Let's call the closest intersection point of the vertical line $(x_1,y_1)$ (note that $x_1=x_0$) and of the horizontal line $(x_2,y_2)$ (note that $y_2=y_0$). Other intersection points can be gotten by simply rotating these around the axes, so you can just ignore them in this problem.

This is the best visual representation I could draw: here.

This is based on a practical problem I have but I think I've included all the relevant information.

Answer 1

Take a look at the following figure:

(where I have assumed that $r_1 < r_2$, that the origin is in the common center, and $a$ the "main angle").

As point $A$ has coordinates $\binom{r_1 \cos a}{r_1 \sin a}$ and $B$ has coordinates $\binom{r_2 \cos b}{r_2 \sin b}$ and they have the same ordinates, we can deduce that :

$$r_1 \sin a=r_2 \sin b \ \iff \ b=\sin^{-1}\left(\frac{r_1}{r_2} \sin a\right)$$

Knowing $b$, you have the coordinates of $B$, and length

$$AB=r_2 \cos b-r_1 \cos a=r_2 \sqrt{1-\left(\frac{r_1}{r_2} \sin a\right)^2} -r_1 \cos a$$

The same kind of reasoning gives the coordinates of point $C$ belonging to the biggest circle and having the same abscissa (this time) as point $A$.